5.
4.2&3: Pumping Lemma
Membership Question
Question: Given regular language L
and string w
how can we check if w L?
Hint: Take a DFA that accepts L
Answer: Take the DFA that accepts L
and check if w is accepted
5
6.
4.2&3: Pumping Lemma
For any w * : For a DFA of L
w w
or
w L w L
Answer: There is a unique path starting
from q0 for w in such DFA.
w L if it ends at a final state.
Theorem 4.5 (p.112) 6
7.
4.2&3: Pumping Lemma
Question: Given regular language L
how can we check
if L is empty: (L )?
Answer: Take the DFA that accepts L
Check if there is a path from
the initial state to a final state
7
9.
4.2&3: Pumping Lemma
Question: Given regular language L
how can we check
if L is finite?
Answer: Take the DFA that accepts L
Check if there is a walk with cycle
from the initial state to a final state
9
10.
4.2&3: Pumping Lemma
DFA
L is infinite
DFA
L is finite
10
11.
4.2&3: Pumping Lemma
Theorem 4.6 (p.112)
There is an algorithm for determining
whether a regular language is empty,
finite, or infinite.
11
12.
4.2&3: Pumping Lemma
Question: Given regular languages L1 and L2
how can we check if L1 L2 ?
Answer: check if( L1 L2 ) ( L1 L2 )
Since L1and L2are regular, (L1 L2 ) (L1 L2 ) is 3
L
also regular. From Thm. 4.6, we can check
whether L3 is empty. If L3= then L1= L2
otherwise L1 L2 .
Theorem 4.7 (p.112) 12
15.
4.2&3: Pumping Lemma
Theorem 4.5 (p.112)
Membership algorithm.
Theorem 4.6 (p.112)
Algorithm for determining whether a regular
language is empty, finite, or infinite.
Theorem 4.7 (p.112)
Algorithm for checking whether 2 regular
languages are the same.
15
16.
Examples
4.2&3: Pumping Lemma
Question: Given regular languages L1 and L2
how can we check if L1 L2 ?
Question: Given regular languages L1 and L2
and any string w
how can we check if w L1 L2 ?
16
17.
Examples cont.
4.2&3: Pumping Lemma
Question: Given regular language L
how can we check if L?
Definition: A language is said to be a
palindrome language if L = LR.
Given regular language L
how can we check if L is a palindrome language.
17
18.
Existence Examples
4.2&3: Pumping Lemma
Question: Given regular language L
how can we check if there exist
any w L such that wR L ?
Question: Given regular language L and string u
how can we check if L contains any
w such that u is a substring of it,
i.e. w =v1u v2 for u, v1,v2 *?
18
19.
Existence Examples cont.
4.2&3: Pumping Lemma
Question: Given regular language L how can
we check if there exist any w L
such that |w| is an odd number ?
Question: Given regular language L how can
we check if L contains an infinite
number of odd-number strings?
You are able to do (4.2) hw# 1~15 except 10 p.113.
19
21.
4.2&3: Pumping Lemma
n n
{a b : n 0}
Non-regular languages
R
{ww : w {a, b}*}
Regular languages
a *b b*c a
b c ( a b) *
etc...
21
22.
4.2&3: Pumping Lemma
How can we prove that a language L
is not regular?
Prove that there is no DFA that accepts L
Problem: this is not easy to prove
Solution: the Pumping Lemma !!!
22
29.
4.2&3: Pumping Lemma
DFA with 4 states
b
b b
q1 a q2 b q3 b q4
a a
29
30.
4.2&3: Pumping Lemma
In walks of strings:
a, aa, aab no state is repeated
b, aba have states repeated
For any string w, |w| 3, then its walk in this
DFA may repeat the state(s) or does not repeat.
b
b b
q1 a q2 a q3 b q4
a a
30
31.
4.2&3: Pumping Lemma
However, aabb
in walks of At least one
bbaa state is repeated
strings with
length 4: abbabb
abbbabbabb...
b
b b
q1 a q2 a q3 b q4
a a
31
32.
4.2&3: Pumping Lemma
Magic
If string w has length | w | 4: number?
The transitions of string w
are more than the states of the DFA
Thus, at least one state must be repeated
b
b b
q1 a q2 a q3 b q4
a a
32
33.
In general, for any DFA:
4.2&3: Pumping Lemma
String w has length number of states
A state q must be repeated in the walk of w
walk of w
...... q ......
Repeated state
33
34.
4.2&3: Pumping Lemma
In other words for a string w:
a transitions are pigeons
q states are pigeonholes
walk of w
...... q ......
Repeated state
34
36.
4.2&3: Pumping Lemma
Take an infinite regular language L
DFA that accepts L
m
states
36
37.
4.2&3: Pumping Lemma
Take string w with w L
There is a walk with label w :
.........
walk w 37
38.
4.2&3: Pumping Lemma
If string w has length | w | m
number of states
of the DFA
then, from the pigeonhole principle:
a state q is repeated in the walk w
...... q ......
walk w 38
39.
4.2&3: Pumping Lemma
Let q be the first state repeated
in the walk
...... q ......
walk w 39
40.
4.2&3: Pumping Lemma
Write w xyz
y
...... q ......
x z 40
41.
4.2&3: Pumping Lemma
Observations: length |x y| m number
of states
length | y| 1 of DFA
y
...... q ......
x z 41
42.
4.2&3: Pumping Lemma
Observation: The string xz
is accepted
y
...... q ......
x z 42
43.
4.2&3: Pumping Lemma
Observation: The string xyyz
is accepted
y
...... q ......
x z 43
44.
4.2&3: Pumping Lemma
Observation: The string xyyyz
is accepted
y
...... q ......
x z 44
45.
i 4.2&3: Pumping Lemma
In General: The string xy z
is accepted i 0, 1, 2, ...
y
...... q ......
x z 45
46.
4.2&3: Pumping Lemma
In General: xy z ∈ L
i
i 0, 1, 2, ...
The original language
y
...... q ......
x z 46
47.
4.2&3: Pumping Lemma
In other words, we described:
The Pumping Lemma !!!
47
48.
4.2&3: Pumping Lemma
The Pumping Lemma
• Given a Infinite Regular language L
• there exists an integer m
• for any string w L with length | w | m
• exists a decompsition of w: wxyz
with |x y| m and | y | 1
• such that: wi L for all i 0, 1, 2, ...
i
xy z
48
49.
4.2&3: Pumping Lemma
Example of pumping lemma
L M = { all strings without
substring 001 }
1 0 0,1
1
q0 q1 0 1 q3
q2
0
49
50.
4.2&3: Pumping Lemma
L M = { all strings without
substring 001 }
1 0 0,1
1
q0 q1 0 1 q3
q2
0
L(M) is infinite and regular, show it
satisfies the pumping lemma.
50
51.
4.2&3: Pumping Lemma
The Pumping Lemma: L is reg. & inf.
• there exists an integer m Let m = 4
• for any string w L with length | w | m
a decomposition w x y z with | x y | m and | y | 1
1 0 0,1
1
q0 q1 0 1 q3
q2
0
• such that: i for all
xy z L i 0, 1, 2, ...
51
52.
4.2&3: Pumping Lemma
For any w L(M), |w| 4:
Then either w= 00...0, or w has 1s;
(1)w contains 0 only: let let x=00, y=0 (or x= , y=0)
and z= rest of the string
(2)w contains one: w is in the form of (1+01)* if w
stops at q0, or w is in the form of (1+01)*0 if w
stops at q1, or w is in the form of (1+01)*000* if w
stops at q2. Since w contains 1, the first 1 in w
must occurs at the first or second symbols of w.
For the first case, let x= , y=1, and z= rest of the
string ; for the second case, let x= , y=01, and z=
rest of the string.
52
53.
4.2&3: Pumping Lemma
Example of pumping lemma
L M = { a nb m : n 2, m 2}
L(M) is infinite and regular, show
it satisfies the pumping lemma.
53
54.
4.2&3: Pumping Lemma
L M = { anbm : n 2, m 2}
The Pumping Lemma: L is reg. & inf.
• there exists an integer m Let m = 5
• for any string w L with length | w | m
a decomposition w x y z with | x y | m and | y | 1
• such that: i for all
xy z L i 0, 1, 2, ...
54
55.
4.2&3: Pumping Lemma
Example of pumping lemma
L M = { uabv : u,v {a,b}* }
L(M) is infinite and regular, show
it satisfies the pumping lemma.
55
56.
4.2&3: Pumping Lemma
L M = { uabv : u,v {a,b}* }
The Pumping Lemma: L is reg. & inf.
• there exists an integer m Let m = 4
• for any string w L with length | w | m
a decomposition w x y z with | x y | m and | y | 1
• such that: i for all
xy z L i 0, 1, 2, ...
56
57.
4.2&3: Pumping Lemma
Discussion
Section 4.2
What is the Pumping lemma?
Show a regular language satisfies
the pumping lemma.
57
59.
4.2&3: Pumping Lemma
Theorem: The language L n n
{a b : n 0}
is not regular
Proof: Use the Pumping Lemma
59
60.
4.2&3: Pumping Lemma
n n
L {a b : n 0}
Since L is infinite
L is regular L satisfies P.L.
(L satisfies P.L.) ( L is regular)
(contrapositive)
Goal is to show L does not satisfy the pumping lemma then L
is not regulatr. by contradiction
i.e. assume L is regular and show L does not satisfy the
pumping lemma contradiction L is not regular
60
61.
4.2&3: Pumping Lemma
(The Pumping Lemma)
• Given a infinite regular language L
for any positive integer m
• there exists an integer m
we pick a string w L
• for any string w L with length | w | m
for any possible decomposition of w
• exist a decompsition of w: w xyz
with |x y| m and | y | 1
some i such that wi= xyiz L
• such that: wi L for all i 0, 1, 2, ...
i
xy z
61
62.
n n 4.2&3: Pumping Lemma
L {a b : n 0}
Since L is infinite, if L is regular then L
satisfies the Pumping lemma. Let m be the
positive integer in the Pumping lemma.
Pick a string w such that: w L
length | w| m
m m
We pick w a b
62
63.
4.2&3: Pumping Lemma
Any decomposition of w = ambm:
m m
w a b xyz
it must be that length | x y | m, | y | 1
m m
m m
xyz a b a...aa...aa...ab...b
x y z
Thus:
k
y a , 1 k m 63
64.
4.2&3: Pumping Lemma
m m
w xyz a b y k
a , 1 k m
i
wi xy z L
k i
a...aa...aa...aa...aa...ab...b...b
x y y ......y z
m k ki m m ( i 1) k m
wi a a b a b
for all i 0, 1, 2, ...
We need to find some i in order to have wi= xyiz L
64
65.
4.2&3: Pumping Lemma
m k ki m m ( i 1) k m
wi a a b a b
for all i 0, 1, 2, ...
We need to find some i in order to have wi= xyiz L
Let i = 2
m ( 2 1) k m m k m
w2 a b a b
Since 1 k m , thus
m k m n n
w2 a b L {a b : n 0}
65
66.
4.2&3: Pumping Lemma
Therefore L does not satisfy the
pumping lemma.
Conclusion: L is not a regular language
66
67.
4.2&3: Pumping Lemma
Non-regular languages R
L {uu : u *}
Regular languages
67
68.
4.2&3: Pumping Lemma
Theorem: The language
R
L {uu : u *} {a, b}
is not regular
Proof: Use the Pumping Lemma
68
69.
4.2&3: Pumping Lemma
R
L {ww : w {a, b}*}
Since L is infinite
L is regular L satisfies P.L.
(L satisfies P.L.) ( L is regular)
(contrapositive)
Goal is to show L does not satisfy the pumping lemma then L
is not regulatr. by contradiction
i.e. assume L is regular and show L does not satisfy the
pumping lemma contradiction L is not regular
69
70.
4.2&3: Pumping Lemma
R
L {uu : u *}
Since L is infinite, if L is regular then L
satisfies the Pumping lemma. Let m be the
positive integer in the Pumping lemma.
Pick a string w such that: w L and
length | w| m
m m m m
We pick w a b b a
70
71.
4.2&3: Pumping Lemma
Any decomposition of w :
m m m m
w a b b a xyz
it must be that length | x y | m, | y | 1
m m m m
xyz a...aa...a...ab...bb...ba...a
x y z
k
Thus: y a , 1 k m 71
72.
4.2&3: Pumping Lemma
m m m m
w xyz a b b a y k
a , 1 k m
i
wi xy z L
2m m
a...aa...aa...aa...a...ab...bb...ba...a
x y y... y z
m k ki m m m m ( i 1) k m m m
wi a a b b a a b b a
for all i = 0, 1, 2, 3, …
We need to find some i in order to have wi= xyiz L
72
73.
4.2&3: Pumping Lemma
m k ki m m m m ( i 1) k m m m
wi a a b b a a b b a
for all i = 0, 1, 2, 3, …
We need to find some i in order to have wi= xyiz L
Let i = 2
m ( 2 1) k m m m m k m m m
w2 a b b a a b b a
Since 1 k m , thus
m k m m m R
w2 a b b a L {uu : u {a, b}*}
73
74.
4.2&3: Pumping Lemma
Therefore: L does not satisfy the
pumping lemma.
Conclusion: L is not a regular language
74
75.
Prove a language is not
regular by pumping lemma
How to write a
good proof?
75
76.
4.2&3: Pumping Lemma
Show L={w: na(w) > nb(w)} is not
regular.
w = ambm-1
Let y = ak
wi L
L is not regular.
76
77.
4.2&3: Pumping Lemma
Non-regular languages
n l n l
L {a b c : n, l 0}
Regular languages
77
78.
4.2&3: Pumping Lemma
Theorem: The language
n l n l
L {a b c : n, l 0}
is not regular
Proof: Use the Pumping Lemma
78
79.
n l n l 4.2&3: Pumping Lemma
L {a b c : n, l 0}
Assume for contradiction
that L is a regular language
Since L is infinite
we can apply the Pumping Lemma
79
80.
n l n l 4.2&3: Pumping Lemma
L {a b c : n, l 0}
Since L is infinite, if L is regular then L
satisfies the Pumping lemma. Let m be the
positive integer in the Pumping lemma.
Pick a string w such that: w L and
length | w| m
We pick w a b c m m 2m
80
81.
Any decomposition of w :
4.2&3: Pumping Lemma
m m 2m
w=a b c xyz
From the Pumping Lemma
it must be that length | x y | m, | y | 1
m m 2m
w = xyz a...aa...aa...ab...bc...cc...c
x y z
k
Thus: y a for some k , 1 k m 81
82.
4.2&3: Pumping Lemma
m m 2m k
w= x yz a b c y a , k 1
i
wi xy z L i 0, 1, 2, ...
= am-k(ak)ibmc2m = am+(i-1)k bmc2m
Let i = 0
0 m k m 2m
w0 x y z a b c L
since (m-k)+m = 2m- k 2m for k 1
82
83.
4.2&3: Pumping Lemma
Therefore: L does not satisfy the
pumping lemma.
Conclusion: L {a nbl c n l : n, l 0}
is not a regular language
83
84.
4.2&3: Pumping Lemma
n l n l
L {a b c : n, l 0} is not regular,
n l k
how about L1 {a b c : k n l} ?
L1 must be non-regular otherwise ….
84
85.
4.2&3: Pumping Lemma
Non-regular languages n!
L {a : n 0}
Regular languages
85
86.
4.2&3: Pumping Lemma
Theorem: The language n!
L {a : n 0}
is not regular
n! 1 2 (n 1) n
Proof: Use the Pumping Lemma
86
87.
4.2&3: Pumping Lemma
n!
L {a : n 0}
Since L is infinite
L is regular L satisfies P.L.
(L satisfies P.L.) ( L is regular)
(contrapositive)
Goal is to show L does not satisfy the pumping lemma
thus L is not regular.
by contradiction 87
88.
4.2&3: Pumping Lemma
n!
L {a : n 0}
Let m be the integer in the Pumping Lemma
Pick a string w such that: w L
length | w| m
m!
We pick w a
88
89.
4.2&3: Pumping Lemma
Any decomposition of w :
m!
w= a xyz
From the Pumping Lemma
it must be that length | x y | m, | y | 1
m m! m
m!
w = xyz a a...aa...aa...aa...aa...a
x y z
k
Thus: y a for some k , 1 k m
89
90.
4.2&3: Pumping Lemma
m! k
xyz a y a , 1 k m
m! k k i m! ( i 1) k
wi a (a ) a
i 0, 1, 2, ...
Let i = 2
2 m! k
w2 xy z a 1 k m
90
91.
4.2&3: Pumping Lemma
2 m! k
w2 xy z a 1 k m
Since: n!
L {a : n 0}
In order to show w2 L, we need to
show m!+k is not a factorial number.
By showing m! < m!+k < ( m+1)! It works only
for all k = 1, 2, 3, …, m when m 2 !
91
92.
However:
4.2&3: Pumping Lemma
m! m! k (m 1)! given m 1
m!< m!+k is trivial
since k 1.
To show m!+k < (m+1)!
From (m+1)! = (m+1) m! = m! + m m!
> m! + k 1 = m!+k
since m k & m! > 1! =1
m! k p! for any p
92
93.
4.2&3: Pumping Lemma
What happen if m = 1?
If m =1, we choose w = a, with x= = z, and
y=a is the only decomposition for w = a.
Then wi = ai for all i = 0, 1, 2, …
Clearly that when i= 3,
w3 = a3 L={an! : n 0}.
Now, put everything together:
93
94.
4.2&3: Pumping Lemma
n!
Show L {a : n 0} is not regular.
Since L is infinite, if L does not satisfy pumping lemma
then L is not regular. So the goal is to prove L does not
satisfy pumping lemma. (it is proved by contradiction)
For any given positive integer m in m pumping lemma,
w L and |w| the
if m =1, we choose w = a, then x = z = , and
y=a is the only decomposition for w = a.
Then wi = ai for all i = 0, 1, 2, …
Clearly that when i= 3,
A contradiction to
w3 = a3 L={an! : n 0}. pumping lemma!
94
95.
4.2&3: Pumping Lemma
n!
Show L {a : n 0} is not regular.
w L and |w| m
If m > 1, we pick w = am!. Consider all possible
decompositions of w = xyz with |xy| m, |y| 1,
k
y must be in the form of y a for some k , 1 k m
m! k k i m! ( i 1) k
wi a (a ) a for i= 0, 1, 2, …
2 m! k
choose i = 2: w2 xy z a 1 k m
From m! < m!+k m! + m < m! + m m! (since m!>1) = (m+1)!
i.e. m! < m!+k < (m+1)!, m!+k is not a factorial number.
w2 L, a contradiction. Thus L does not satisfy
pumping lemma, and therefore L is not regular.
95
96.
4.2&3: Pumping Lemma
Non-regular languages L n l k
{a b c : k n l}
Regular languages
96
97.
4.2&3: Pumping Lemma
Theorem: The language L {a nbl c k : k n l}
is not regular
Proof: Use the Pumping Lemma
Choose a w from L and, later,exist an i such that wi L
What does wi L imply?
97
98.
4.2&3: Pumping Lemma
n l k
L {a b c : k n l}
Since L is infinite, if L is regular then L
satisfies the Pumping lemma. Let m be the
positive integer in the Pumping lemma.
Pick a string w such that: w L
length | w| m
We pick
99
99.
4.2&3: Pumping Lemma
Any decomposition of w :
m! ( m 1)!
w=a c xyz
From the Pumping Lemma
it must be that length | x y | m, | y | 1
m! (m 1)!
m! ( m 1)!
w = xyz a c a...aa...aa...ac...cc...c
x y z
Thus: y a k
for some k , 1 k m
100
100.
4.2&3: Pumping Lemma
m! ( m 1)! k
xyz a c y a , 1 k m
m! k k i ( m 1)! m! ( i 1) k ( m 1)!
wi a (a ) c a c
i 0, 1, 2, ...
How to choose i such that wi L?
i.e. choose i such that m!+(i-1)k = (m+1)!
m! m Is such i an integer?
choose i 1
k
101
101.
4.2&3: Pumping Lemma
m! ( m 1)! k
xyz a c y a , 1 k m
m! k k i ( m 1)! m! ( i 1) k ( m 1)!
wi a (a ) c a c
i 0, 1, 2, ...
Since 1 k m, k | m!
m! m
when i 1 (: a positive integer)
k
m!+(i-1)k = (m+1)! which causes wi L
A contradiction to pumping lemma!
102
102.
4.2&3: Pumping Lemma
Therefore: L does not satisfy the
pumping lemma.
Conclusion: L {a b c : k
n l k
n l}
is not a regular language
Theorem: The language L {a nbl c k : k n l}
is not regular 103
103.
4.2&3: Pumping Lemma
n! 2 6 24 120
Show {a : n 0} {a, a , a , a , a , ...} not regular.
Since L is infinite, if L does not satisfy pumping lemma
then L is not regular. So the goal is to prove L does not
satisfy pumping lemma. (it is proved by contradiction)
For any given positive integer m in m pumping lemma,
w L and |w| the
if m =1, we choose w = a, then x = z = , and
y=a is the only decomposition for w = a.
Then wi = ai for all i = 0, 1, 2, …
Clearly that when i= 3,
A contradiction to
w3 = a3 L={an! : n 0}. pumping lemma!
104
104.
4.2&3: Pumping Lemma
n! 2 6 24 120
L {a : n 0} {a, a , a , a , a , ...}
w L and |w| m
If m > 1, we pick w = am!. Consider all possible
decompositions of w = xyz with |xy| m, |y| 1,
k
y must be in the form of y a for some k , 1 k m
m! k k i m! ( i 1) k
wi a (a ) a for i= 0, 1, 2, …
2 m! k
choose i = 2: w2 xy z a 1 k m
From m! < m!+k m! + m < m! + m m! (since m!>1) = (m+1)!
i.e. m! < m!+k < (m+1)!, m!+k is not a factorial number.
w2 L, a contradiction. Thus L does not satisfy
pumping lemma, and therefore L is not regular.
105
105.
4.2&3: Pumping Lemma
Consider the following two languages,
determine which is regular.
n l
L1 {a b : n l}
n l
L2 {a b : n l}
Both are non-regular.
Which is easier to prove by pumping lemma?
106
106.
4.2&3: Pumping Lemma
Consider the following two languages,
determine which is regular.
n l
L1 {a b : n l}
n l
L2 {a b : n l}
Both are non-regular.
Which is easier to prove by pumping lemma?
107
107.
4.2&3: Pumping Lemma
Non-regular languages n l
L {a b : n l}
Regular languages
108
108.
4.2&3: Pumping Lemma
Theorem: The language L n l
{a b : n l}
is not regular
Proof: Use the Pumping Lemma
The proof is similar to the proof in
n l k
L {a b c : k n l}
109
109.
4.2&3: Pumping Lemma
n l
L {a b : n l}
Since L is infinite
L is regular L satisfies P.L.
(L satisfies P.L.) ( L is regular)
(contrapositive)
Goal is to show L does not satisfy the pumping lemma
thus L is not regular.
by contradiction 110
110.
4.2&3: Pumping Lemma
n l
L {a b : n l}
Let m be the integer in the Pumping Lemma
Pick a string w such that: w L
length | w| m
We pick w a b m! ( m 1)!
111
111.
4.2&3: Pumping Lemma
Any decomposition of w :
m! ( m 1)!
w=a b xyz
From the Pumping Lemma
it must be that length | x y | m, | y | 1
m! (m 1)!
m! m 1!
w = xyz a b a...aa...aa...ab...bb...b
x y z
Thus: y a k
for some k , 1 k m
112
112.
4.2&3: Pumping Lemma
m! ( m 1)! k
xyz a b y a , 1 k m
m! k k i ( m 1)! m! ( i 1) k ( m 1)!
wi a (a ) b a b
i 0, 1, 2, ...
How to choose i such that wi L?
i.e. choose i such that m!+(i-1)k = (m+1)!
m! m Is such i an integer?
choose i 1
k
113
113.
4.2&3: Pumping Lemma
m! ( m 1)! k
xyz a b y a , 1 k m
m! k k i ( m 1)! m! ( i 1) k ( m 1)!
wi a (a ) b a b
i 0, 1, 2, ...
Since 1 k m, k | m!
m! m
when i 1 (: a positive integer)
k
m!+(i-1)k = (m+1)! which causes wi L
A contradiction to pumping lemma!
114
114.
4.2&3: Pumping Lemma
Therefore: L does not satisfy the
pumping lemma.
Conclusion: L n l
{a b : n l}
is not a regular language
Theorem: The language L n l
{a b : n l}
is not regular
115
115.
4.2&3: Pumping Lemma
Theorem: The language L n l
{a b : n l}
is not regular
Proof:
Another way to prove L is not regular by
closure property of regular family.
L is regular then L is also regular
What is the complement of L?
116
116.
4.2&3: Pumping Lemma
is not regular
n l
L {a b : n l}
Proof:
Assume L is regular then L is also regular
Let R = L L(a*b*)
an intersection of two regular languages,
thus R is a regular language too.
In fact, R = {anbn : n 0} which is not regular.
A contradiction!
Therefore, the assumption that L is regular is
not true, i.e., L is not regular.
117
117.
Similar problem:
4.2&3: Pumping Lemma
Show L {a b c : k n l} is not regular.
n l k
Show L ={ w: na(w) nb(w) } is not regular.
Is L* regular or non-regular?
Prove or disprove:
If L1 and L2 are non-regular, then L1 L2 must be non-regular.
If L1 and L2 are non-regular, then L1 L2 must be non-regular.
If L is regular then any subset of L must be regular too.
If L is non-regular then any subset of L must be non-regular .
118
118.
4.2&3: Pumping Lemma
Prove or disprove:
If L1 L2 and L1 are regular, then L2 must be regular.
If L1 L2 is regular and L1is finite, then L2 must be
regular.
The family of regular languages closed under finite number
union.
The family of regular languages closed under infinite number
union.
119
119.
4.2&3: Pumping Lemma
Regular or Non-regular?
R
L1 {uww v : u, v, w {a, b} }.
R
L2 {uww v : u, v, w {a, b} , | u | | v |}.
// L1 is regular, but L2 is non-regular.
120
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