1. Closure Properties of
Regular Languages
This is the partial material
from Section 4.1, 3.1 & 3.2
1

2. For regular languages L1 and L2
we will prove that:
Union: L1 L2
Concatenation: L1L2
Star: are regular
L1 *
Languages
Reversal: L1R
Complement: L1
Intersection: L1 L2 2

3. We say: Regular languages are closed under
Union: L1 L2
Concatenation: L1L2
Star: L1 *
Reversal: L1R
Complement: L1
Intersection: L1 L2 3

4. Regular language L1 Regular language L2
L M1 L1 L M2 L2
NFA M1 NFA M2
Single final state Single final state
4

5. Example
M1
a
n
L1 {a b} b
n 0
M2
L2 ba b a
Both with Single final state 5

6. Union
Make it into
NFA for L1 L2 non-final
M1
New final
state
M2
An NFA for L1 L2 with Single final state.
6

7. Example
n
NFA for L1 L2 {a b} {ba}
n
L1 {a b}
a
b
L2 {ba}
b a
7

8. Concatenation
Make it into
NFA for L1L2 non-final
M1 M2
A new
final state
8

9. Example
n n
NFA for L1L2 {a b}{ba} {a bba}
n
L1 {a b}
a L2 {ba}
b b a
9

10. Star Operation
NFA for L1 *
A new edge
L1 *
M1
A new
initial state A new
final state
FA for r.e.
A new edge
10

11. Example
NFA for n
L1* {a b} *
Become
n non-final
L1 {a b}
a
b A new
final state
11

12. Reverse
R
NFA for L1
L1 M1 M1
1. Reverse all transitions
2. Make initial state final state
and vice versa
12

13. Example
M1
n
a
L1 {a b : n 0}
b
M1
R n
a
L1 {ba : n 0}
b
Cont. back to Intersection
13

14. Complement
L1 M1 L1 M1
IMPORTANT!!
1. Take the DFA that accepts L1
2. Make final states non-final,
and vice-versa
14

15. Example
M1
a a, b
n
L1 {a b} b a, b DFA！
M1
a a, b
n
L1 {a, b} * {a b}
b a, b
How to do it if an NFA is given instead? 投影片 13
15

16. Intersection
DeMorgan’s Law: L1 L2 L1 L2
L1 , L2 regular
L1 , L2 regular
L1 L2 regular
L1 L2 regular
L1 L2 regular
16

17. Example
L1 {a b} regular
n
L1 L2 {ab}
L2 {ab, ba} regular regular
The cor of two languages is
cor ( L1 , L2 ) {w : w L1 or w L2 }
Show that the family of regular languages is closed
under the cor operation.
17

18. A Construction way for L1 L2
disjoint same
L1 L( M 1 ) with DFA M 1 (Q, , 1 , q0 , F1 )
L2 L( M 2 ) with DFA M 2 ( P, , 2 , p0 , F2 )
ˆ
An automaton M (Q P, , ˆ, (q0 , p0 ), F ) will
ˆ
be constructed for L1 L2 such that
for any ( qi , p j ) Q P
ˆ((q , p ), a ) (q k , pl ) iff ( qi , a ) q k and ( p j , a) pl
i j 1 2
and
ˆ
(qi , p j ) F only if qi F1 and p j F2
18

19. We use MI to denote the automaton constructed
above for L1 L2, and so are I, FI.
w L (MI ) I( (q0,p0), w)= (qf,pf) FI qf F1
and pf F2
let w = a1a2…ak, since I( (q0,p0), w)= (qf,pf) FI ,
there exists a walk from (q0,p0) to (qf,pf) with label
w.
say (q0,p0) (qi,pj) (qr,ps) … (qf,pf) there
exists a walk from q0 to qf with label w, and there
exists a walk from p0 to pf with label w. Since qf
F1 and pf F2 w L1 L 2
19

20. Find a FA for L1 L2
from machines of L1 and L2
n n
L1 {a : n 1} {ba : n 0} and
n
L2 {b} {ba : n 2}
Use your words to explain why
ˆ
L( M ) L1 L2
20

21. Example
• If L is a regular language, prove that the
language {uv: u L, v LR} is also regular.
21

22. Regular Expressions
22

23. Regular Expressions
Regular expressions
describe regular languages
The symbol of concatenation
can by omitted, i.e. ( a + bc )*
Example: (a b c) *
describes the language
a, bc * , a, bc, aa, abc, bca,...
23

24. Recursive Definition p.72
Primitive regular expressions: , , a
Given regular expressions r and
1 r2
r1 r2 The symbol of concatenation can be
omitted, so can be written as r1 r2
r1 r2
Are regular expressions
r1 *
r1
A string is a regular expression it can be derived from
primitive r.e.’s by a finite # of applications of the rules above.
24

25. Examples
regular expression or Not regular expression:
a (X)
a* b*
a b * (c ) a* c b c
3 2 m 2 n
a ( b b )a * a ( b b )a
Precedence rules:
25

26. Languages of Regular Expressions
Lr : language of regular expression r
Example
L ( a b c) * , a, bc, aa, abc, bca,...
26

27. Definition
For primitive regular expressions:
L
L
La a
27

28. Definition (continued)
For regular expressions r and
1 r2
L r1 r2 L r1 L r2
L r1 r2 L r1 L r2
L r1 * L r1 *
L r1 L r1
28

29. Example of Language L(r)
• Regular expression r a b *
Lr L((a b)*) L(a b) *
( L(a) L(b)) *
({a} {b})* {a, b}*
29

30. Example
• Regular expression: a b a*
L a b a* L a b L a*
L a b L a*
La Lb La *
a b a *
a, b , a, aa, aaa,...
a, aa, aaa,..., b, ba, baa,...
30

31. Example
Regular expression r a b * a bb
Lr a, bb, aa, abb, ba, bbb,...
{w {a, b} : w ends with a or bb}
Find a r.e. for L={w {a,b}*: w has aa as a substring}
31

32. Example
Regular expression r aa * bb * b
2n 2m 1
Lr {a b : n, m 0}
Describe L(r) if r = (aa)*b(aa)* + a(aa)*ba(aa)*.
Find a r.e. for L={ambn: m+n is even}
Which is correct:
{a2nba2n: n 0} {a2n+1ba2n+1: n 0}; {a2nba2m:
m,n 0} {a2n+1ba2m+1: m,n 0}; {anbam:
m,n 0}
32

33. Example of L(r)
Regular expression a , ( *)*
33

34. Example
Find a regular expression for
L(r ) = { all strings with at least
two consecutive 0 }
Regular expression r (0 1) * 00 (0 1) *
Find a regular expression for
L(r ) = { all strings with exactly
two consecutive 0 }
34

35. Example
Find a regular expression for
L(r ) = { all strings without
two consecutive 0 }
Case1: no 0, i.e. 1* ; or
case2: has 0, such like(0, 01, 10, 101, 011, 110,
010,0111, 1011, 1101, 1110, 0101, 0110, 1010, 0101,…)
--0--, (1+01)*0(1+10)*
Possible answers are: 1* + (1+01)*0(1+10)*;
(1+01)*(0+ )(1+10)*; (1+01)*(0+ ); (0+ )(1+10)*;
35

36. Example
L = { all strings without two consecutive 0 }
r1 = 1* + (1+01)*0(1+10)*
r2 (1 01) * (0 )
r3 (0 )(1 10 ) *
L(r1 ) L(r2 ) L(r3 )
are equivalent
r1 , r2 r
regular expr.
(1*011*)*(0+ )+1*(0+ ) is also an equivalent r.e.
36

37. Equivalent Regular Expressions
Definition:
Regular expressions r1 r2 , i.e. they
are equivalent if L(r1) L(r2 )
Try Hw #20 on p. 77
37

38. Example
Find regular expressions for
n m
L1 {a b : n 4, m 3}
The complement of L1
38

39. Example
Find a regular expression that denotes all bit strings
whose value, with leading bit 1, when interpreted as
a binary integer, is greater than or equal to 20.
Find a r.e. r such that
L(r) = {w {0, 1}*: w = 1v and binary(w) 20}
Homework
(p.108) 2, 3, 5 ~8, 12, 13, 14, hand in: 2b, 13
(p.75) 1~21, 26~28 hand in: 6bd, 14, 16ab, 17b, 18b, 20bd,
21, 28
39

40. Regular Expressions
and
Regular Languages
40

41. Theorem
Languages
Regular
Generated by
Languages
Regular Expressions
41

42. Theorem - Part 1
Languages
Regular
Generated by
Languages
Regular Expressions
1. For any regular expression r
the language L(r ) is regular
42

43. Theorem - Part 2
Languages
Regular
Generated by
Languages
Regular Expressions
2. For any regular language L there is
a regular expression r with L( r ) L
43

44. Proof - Part 1
1. For any regular expression r
the language L(r ) is regular
Proof by induction on the size of r
44

45. Induction Basis
Primitive Regular Expressions: , , a
NFAs
L ( M1 ) L( )
regular
L( M 2 ) { } L( )
languages
a
L( M 3 ) {a} L(a )
45

46. Inductive Hypothesis
Assume for regular expressions r1 and r2 that
L(r1) and L(r2) are regular languages
We will prove:
L r1 r2
L r1 r2 Are regular
Languages
L r1 *
L r1
46

47. By definition of regular expressions:
L r1 r2 L r1 L r2
L r1 r2 L r1 L r2
L r1 * L r1 *
L r1 L r1
47

48. By inductive hypothesis we know:
L(r1) and L(r2 ) are regular languages
We also know:
Regular languages are closed under:
Union L r1 L r2
Concatenation L r L r2
1
Star L r1 *
48

49. Therefore:
L r1 r2 L r1 L r2
Are regular
L r1 r2 L r1 L r2
languages
L r1 * L r1 *
And trivially
L((r )) is a regular language
1
49

50. Example
Find an NFA for L(r) if r = (a+bb)*(ba*+ )
a a
b b b b
r = (a+bb)* a
b b
50

51. Example (Cont’d)
Find an NFA for L(r) if r = (a+bb)*(ba*+ )
a
r1 = (a+bb)*
b b
a
It is necessary!
r2 = (ba*+ ) b
51

52. Example
Find an NFA for L(r) if r = (a*b)*
Construct FAs for the expressions
( ) and a . What are the languages
that they represent?
52

53. Proof – Part 2
2. For any regular language L there is
a regular expression r with L( r ) L
Proof by construction of regular expression
53

54. Since L is regular take the
NFA M that accepts it
L( M ) L
Single final state
54

55. From M construct the equivalent
Generalized Transition Graph
in which transition labels are regular
expressions
Example:
M
a c a c
a, b a b
55

56. Another Example: b b
a
q0 q1 a, b q2
b
b b
a
q0 q1 a b q2
b
56

57. Reducing the state q1 : b b
a
q0 q1 a b q2
b
b b Add edges from qi to qj if there
was edges qi to qj via q1, i, j 1:
a
q0 q1 a b q2 q0 to q0 via q1:
q0 to q2 via q1:
q2 to q0 via q1:
b
bb * a q2 to q2 via q1:
b
q0 bb * (a b) q2
57

58. Resulting Regular Expression:
bb* a b
q0 bb * (a b) q2
r (bb * a ) * bb * (a b)b *
L(r ) L( M ) L
58

59. In General
For any transition graph satisfying
q0 F and | F = {qf} | = 1
then for any state q , q q0 , q qf
will be removed.
59

60. a d
Removing state q :
c
q1 q2
Add edges from qi to qj if b
there were edges qi to qj via q:
e f h i
q
g
60

61. a
a d
d
Removing state q :
c
q1 q2
b
e f h i
q1 to q1 via q:
q
g
a +eg*f d
c
q1 q2
b 61

62. a d
Removing state q : c
q1 q2
b
e f h i
q1 to q1 via q:
to q2 via q: q
g
a+eg*f d
c
q1 q2
b+eg*i 62

63. a d
Removing state q :
c
q1 q2
b
e f h i
q1 to q1 via q:
to q2 via q: q
q2 to q1 via q:
g
a+eg*f d
c+hg*f
q1 q2
b+eg*i 63

64. a d
Removing state q :
c
q1 q2
b
e f h i
q1 to q1 via q:
to q2 via q: q
q2 to q1 via q:
to q2 via q: g
a+eg*f d+hg*i
c+hg*f
q1 q2
b+eg*i 64

65. The final transition graph:
r4
q0 qf
r2
The resulting regular expression:
r r2 r4 *
L(r ) L( M ) L
65

66. The final transition graph:
r4
r3
q0 qf
r2
The resulting regular expression:
r r2 (r4 r3r2 ) *
L(r ) L( M ) L
66

67. The final transition graph:
r1 r4
r3
q0 qf
r2
The resulting regular expression:
r r1 * r2 (r4 r3r1 * r2 ) *
L(r ) L( M ) L
67

68. Removing state q :
e
c
if qi q qj
a b
e
d c
if qi q qj
a
68

69. Removing state q : e
f q g
d c
Add edges from qi to qj if there a b
were edges qi to qj via q: q1 h q2
i
r1 r4 j k
l m n o
r2
q1 q2
r3 q3
r5 q1 to q1 via q: p
r6 r7 r8
to q2 via q:
q3 to q3 via q: q3 to q1 via q:
q2 to q1 via q: to q2 via q:
r9 to q2 via q: to q3 via q:
to q3 via q:
69

70. Last Words about
Conversion from NFA to Reg. Exp.
• Remove any inaccessible states (from q0).
• If the initial state is also a final state…
• If there is more than one final states…
Hw for 3.2 (p.87): 1 ~13, 18.
70

71. L is regular
a reg. exp. r L=L(r)
a FA M L=L(M)
Show L is regular by finding its regular
expression:
L = {w {a,b}*: na(w) is even, nb(w) is odd}
It is difficult to find its reg. exp. directly.
Use FA with states:
eo representing so far it has even # of as and
odd # of bs being read, etc..
71

72. Closure Properties of
Regular Languages
Remaining material of
Section 4.1
72

73. Closure Properties
• Closure Properties of Regular Languages
– Closure under Simple Set Operations
• union, intersection, concatenation,
complementary, star-closure, reversal
– Closure under Other Operations
• Homomorphism (Def. 4.1, p.103) skip
• Right Quotient (Def. 4.2, p. 104)
73

74. Right Quotient L1/ L2
• The right quotient of L1 with L2 is
L1/L2 = {x: xy L1 for some y L2}
L1={ab, abb, abbb}, L2 ={bman: m > n}
L1 / L2 = ?
Thm 4.4 (p.106) The family of regular languages
is closed under right quotient with a regular
language.
74

75. Right Quotient L1/ L2
• The right quotient of L1 with L2 is
L1/L2 = {x: xy L1 for some y L2}
Let L1=L(a*baa*) and L2 =L (ab* )
Find L1/L2 & L2/L1
Thm 4.3 (p.104) & 4.4 (p.106) The family of regular
languages is closed under arbitrary
homomorphisms and right quotient with a
regular language. 75

76. Right quotient example (p.109, Hw #11)
• Give an example of L1 & L2 such that
L1 = L1L2/L2.
• Give an example to show that
L1 = L1L2/L2 is not true.
Do hw (p.109) : 10, 11, 15 of your own.
76