

Chapter 4
Properties of
Regular Languages

1


4.1: Closure Properties of RLs (1) 


A language L over an alphabet S is regular
if it
 has a regular set (expression) over S

is accepted y DFA, NFA, or NFA-e

is generated by regular grammar
2



4.1: Closure Properties of RLs (2) 
3


4.1: Closure Properties of RLs (3) 

If L1 and L2 are regular languages, then
 L1  L2
Union
 = {w * : w  L1 or w  L2}



L1  L2
Intersection

L1 L2
Concatenation

L*
Kleene Star

L1
Complement
 = {w * : w  L1 and w  L2}
 n languages: L1L2 …Ln = {w1w2…wn : wi  Li}
 one languages: Ln = {w1w2…wn : wi  L}
 L*= L0  L1  L2  …
 L = * - L = {w : w * and w  L}
are also regular languages
4


4.1: Closure Properties of RLs (4) 





Prove that RLs are closed under complementation.
If L is a regular language over S, then L = S* - L is
regular language.
Proof:
 If L is regular, there exists a DFA M
recognizing L.
 We can construct a DFA M’ for L by copying
M to M’ except that all final states in M are
changed to non-final, and all non-final states to
final.
See next slide for a formal proof
5


4.1: Closure Properties of RLs (5) 








Let M1=(Q, S, , q0, F) a DFA that accepts L(M1)
Let M2=(Q, S, , q0, Q-F) a DFA that accepts L(M2)
Obviously both languages are regular languages
Per definition 2.2: L(M) = {w * : (q0, w)  F}
The following are both true
 w * : (q0, w)  F
 (q0, w)  Q – F
 w * : (q0, w)  Q - F
 (q0, w)  F
Thus L(M2) = L(M1)
L(M1) is an arbitrary regular language and its
complement is also a regular language, therefore the
regular languages are closed under complementation
6


4.1: Closure Properties of RLs (6) 



Are regular languages closed under intersection?
If L1 and L2 are RLs, then L1  L2 is RL.
Proof:
 Since RLs are closed under union and
complementation, they are also closed under
intersection
 L1, L2 are RLs, so L1, L2 are RLs, and L1  L2
is RL, so L1  L2 is RL.


Thus L1  L2 = L1  L2 is RL.
7


4.1: Closure Properties of RLs (7) 


Are regular languages closed under difference?
If L1 and L2 are RLs, is L1 - L2 RL? Why?




L1 - L2 = L1  L2
Since RLs are closed under intersection and
complementation, they are also closed under
difference.
8



4.1: Closure Properties of RLs (8) 

Let L, M and N are regular expressions, which of
the followings are correct?

Commutative Rules:
 LM=ML
 LM = ML

Associative Rules:
 (L  M)  N = L  (M  N)
 (LM)N = L(MN)
9



4.1: Closure Properties of RLs (9) 

Which of the followings are correct?

Left Distributive Rules:
 L(M  N) = LM  LN
 L  (MN) = (L  M)(L  N)

Right Distributive Rules:
 (M  N)L = ML  NL
 (MN)  L = (M  L)(N  L)
10


4.1: Closure Properties of RLs (10) 

Identities:

 is the identity for union:
L=L=L

e is the identity for concatenation:
 eL = Le = L

11


4.1: Closure Properties of RLs (11) 

Other Rules for Kleene Closure






(L*)* = L*
L+ = LL* = L*L
L* = L+ | e
*=e
e* = e
12


4.2: Elementary Questions about RLs 

Membership question:
 If L is a RL over S, then for any string w  S*
there exists an algorithm for determining
whether or not w is in L.
 DFA

Empty, finite, or infinite question:
 There exists an algorithm for determining
whether a RL is empty, finite, or infinite.
 DFA (path from start to final, cycle)


Equality question:
 If L1 and L2 are RLs, then there exists an
algorithm to determine whether or not L1 = L2.
 L3 = (L1  L2)  (L1  L2) =  iff L1 = L2
13


4.3: Non-regular Languages (1)

The language
 L = {an bn | n  0} is not regular
 There is NO DFA/NFA that accepts L

Theorem:
If L1 is RL and L2 is not-regular, then
 The language L = L1  L2 is not necessarily
regular
Proof:
 L1 = a*b* is regular
 L2 = {an bn | n  0} is not regular
 L1  L2 = L2 which is not regular



14



4.3: Non-regular Languages (2)




The language
 L = {an bm | n, m  0 and n  m} is not regular.
Proof:
 L is regular iff L  a*b* is regular.

But L  a*b* = {an bn | n  0}, which we know
is not regular.

Then L is not regular.
15


4.3: Non-regular Languages (3)



Many languages are non-regular:
 {anbn | n  0}
 {wwR} | w  * }
2
 {an | n  0}
 {ap | p is prime}
 set of well-formed parentheses
 set of palindromes: {w = wR}
 …...
16


4.3: Non-regular Languages (4)


We want to prove that L = {akbk | k  0} is nonregular. Prove by contradiction:


Assume that there is a DFA M which
recognizes L. Let n be the no. of states in M
Consider the acceptance of the input arbr where
r  n:
q0 a q1 a q2 a
…...
a qr b qr+1 b
ar
…... b q2r
br
path for accepting arbr

17


4.3: Non-regular Languages (5)


Since r  n and M has only n states, there must
be at least one state visited twice in the first r
transitions. Let this state be visited at the ith and
the jth steps, where j  i.
qi = qj
qf  F
q0



a path in M
By skipping the loop, ar-(j-i)br should also be
accepted by M,
but this is contradictory since ar-(j-i)br  L
18


4.3: Non-regular Languages (6)


2
k
{a
We want to prove that L =
| k  0} is nonregular. Prove by contradiction:


Assume that there is a DFA M which
recognizes L. Let n be the no. of states in M
M should also accept the string an2
q0 a q1 a q2 a
…...
a q2
n
path for accepting an2

19


4.3: Non-regular Languages (7)


Since n2  n and M has only n states, there must
be at least two equal states from q0 to qn2. Let
them be qi and qj where j - i = m  n.
A loop of
length m
qi = qj
a path in M
qn2  F
q0


By repeating the loop one more time, a(n2+m) is
also accepted by M, which is a contradiction,
since (n2+m) cannot be a square (the next
square after n2 is (n+1)2 but n2+m  (n+1)2)
20


4.3: Non-regular Languages (8)


We want to prove that L = {ap | p is a prime} is
non-regular. Prove by contradiction:


Assume that there is a DFA M which
recognizes L. Let n be the no. of states in M
From Number Theory, we know that the no. of
primes are infinite, so there exists a prime p  n
q0 a q1 a q2 a
…...
a qp
path for accepting ap

21


4.3: Non-regular Languages (9)

Since p  n and M has only n states, there must
be at least two equal states from q0 to qp. Let
them be qi and qj where j - i = m  0
qi = qj
A loop of
length m
a path in M
q0



qp  F
By repeating the loop (p-m) times, a(p-m)m + (p-m)
= a(p-m)(m+1) is also accepted by M, which is a
contradiction since (p-m)(m+1) is not a prime,
since it can be divided by (p-m) and (m+1)  1
22


4.3: Pumping Lemma (1)


If L is a regular language, there is a DFA which
recognizes L.
 All finite languages are regular
 , {a}, {abaabb, b, baabbbbb, bb}


So, only infinite languages are interesting for
this discussion.

What are non-regular languages?

Pumping Lemma
 A technique for proving a language L is NOT
regular
23


4.3: Pumping Lemma (2)



Any given DFA has a finite number of states
Let k = |Q|
Consider any accepted input w = a1…an, such that
|w| = n  k
 DFA follows a path of states q0…qn:
q0




a1
q1
a2
…
an-1
qn-1
an
qn
This state numbering has nothing to do with
state names in Q
May repeat states in this path
24


4.3: Pumping Lemma (3)


Since |w|  k, path must contain at least one loop
 Consider any one such loop (cycle)
 Break string & path into three parts: before,
during, & after loop
…
aj
q0


a1
…
ai
ai+1
qi=qj
aj+1
…
an
qn
Given other inputs, could loop an arbitrary number
of times
 (a1…ai)(ai+1…aj)*(aj+1…an) also accepted
25


4.3: Pumping Lemma (4)





If language L is regular, then for any w  L that is
longer than |Q|, there is a substring of w that can be
repeated (pumped) any number of times with
resulting string remaining in the language
Lemma:
 Let G be the state diagram of a DFA with k states.
Any path of length k in G contains a cycle.
Proof:
 A path of length k contains k+1 nodes. Since there
are only k nodes (states) in G, there must be a
node, call it qi, that occurs in at least two positions
in the path. The subpath from the first occurrence
of qi to the second produces the desired cycle.
26


4.3: Pumping Lemma (5)

Let G be the state diagram of a DFA with k
states and let p be a path of length  k.




The path p can be subdivided into
subpaths u, v, and z where p = uvz
The length of uv is |uv|  k, and v is a
cycle
27


4.3: Pumping Lemma (6)




Pumping Lemma
Let L be a regular language is accepted by a DFA
M with k states. Let for ALL w  L and |w|  k.
Then w can be divided into three parts, w = uvz,
satisfying the following conditions:

|uv|  k,

|v|  1, and

for all i  0, uviz  L
28


4.3: Pumping Lemma (7)




Proof:
Let w  L with |w| = n  k
where w(i) is the ith symbol of the string w.
q0 w(1) q1 w(2) q2 w(3) …... w(n) qn


Since n  k and M has only k states, there must be
at least one repeated state from q0 to qn
Let qi be the first such repeated state
qi

q0
a path in M
29
qn  F


4.3: Pumping Lemma (8)

qi
q0




a path in M
qn  F
Let u be the string obtained by traversing from
q0 to qi, v be the string obtained by traversing
the loop once (so |v|  1)
In the traversal from q0 to qi and then through
the loop once back to qi, nothing except qi
repeats, thus |uv|  k
By traversing the loop 0 or more times, we
obtain uviz for all i  0 and they should all be
accepted by the DFA, i.e., in L.
30



4.3: Pumping Lemma (9)

How to use pumping lemma to prove some L isn’t
regular?
 Lemma describes a property of regular
languages.
 Use proof by contradiction: If L doesn't have
this property, it is NOT regular

We cannot use the pumping lemma to prove a
language is regular
 Since we have to check every string in L
31



4.3: Pumping Lemma (10)










We assume the given language L is regular
Let n be the constant of pumping lemma
Find a string w  L where |w|  n
Divide w into 3 parts, w = uvz, where
 |uv|  n
 |v|  1
Claim that uviz are also in L for all i
Find one i such that uviz  L
Contradiction
Only one string w is needed
 do not need to consider all general cases
 choose w that can led to contradiction easily
32


4.3: Pumping Lemma (11)


Example 1:
 L = {am bm | m  0}
 Assume L is regular, n is the # of DFA states
 Let w = anbn
where |w| = 2n  n
where p + j  n and j  1
 w = ap aj an-p-j bn
w=u v z


u & v contain only a’s, since |uv|  n
Pick i = 2:
 uv2z = ap aj aj an-p-j bn = an aj bn
 uv2z contains more a’s than b’s, since |v|  1
 uv2z  L


Contradiction and hence L is not regular
33


4.3: Pumping Lemma (12)

Example 2:
 L = {w | w  {a, b}*, w has equal number of a’s
and b’s}
 w = an bn
 Same argument as previous example

Choice of w is important
 Different w’s may lead to different proofs, or
none being possible
Choice of i is important.
 Different i’s may lead to different proofs, or
none being possible



34


4.3: Pumping Lemma (13)


Example 3:
 L = {(ab)i | i  0}
 L is regular: L = (ab)*
 i.e., for any w, there is some split u, v, z that allows
pumping

35


4.3: Pumping Lemma (14)





Example 4:
Prove that the language L of well-formed
parentheses is non-regular
Assume L is regular and let n be the constant of
pumping lemma
Let w = (n)n  L, |w| = 2n  n
 we do not need to consider other possible cases
like ()()
 we only need to ensure |w|  n
Then w = (m (j (n-m-j )n where j  1 and m + j  n
u


v
z
36


4.3: Pumping Lemma (15)


By pumping lemma,
(m(ji(n-m-j )n = (n+j(i-1) )n  L for all i


When i = 0
 (n-j )n  L
 since j  1  n-j  n

Contradiction 

We can choose i = 2, i = 3, ...
 but cannot choose i = 1
L is not regular
37


4.3: Pumping Lemma (16)

To prove L is regular

1. Constructing a FSA (NFA or DFA), RE, or RG for L

2. Using closure property of regular sets
 Let ⊕ be a binary operation (such as union,
concatenation and intersection)
L3 = L1 ⊕ L2 is regular, if both L1 and L2 are regular


Let ⊡ be an unary operation (such as complementation
and Kleene star)
 L1 = ⊡ L2 is regular, if L2 is regular
38


4.3: Pumping Lemma (17)

Example
 Prove that L is regular, where L is the set of all
strings with three consecutive 0's and S = {0,1}
 Let
L1 = (0 | 1)*, which is regular and
L2 = 000, which is also regular
 then L = L1．L2．L1
 Hence L is regular due to the closure property
of regular sets under concatenation.

39


4.3: Pumping Lemma (18)

Prove L is not regular
 1. Using Pumping Lemma
 2. Using closure property of regular sets
 Let ⊕ be a binary operation (such as union,
concatenation and intersection)
if L3 = L1 ⊕ L2 and L1 is regular and L3 is not regular,
then L2 must be non-regular
⊡ be an unary operation (such as
complementation and Kleene star)
 Let
if L1 = ⊡ L2 and L2 is not regular, then L1 is nonregular

40


4.3: Pumping Lemma (19)


L = {w  {a, b}*, w has equal number of a’s and
b’s}
 Consider
 By
L’ = L  a*b* = {an bn | n  0}
closure properties of RLs:
if L is regular, then L’ is also regular
But since L’ not regular, so L can’t be regular

41


4.3: Pumping Lemma (20)

Example
 Prove that L is not regular, where L is the set of all
strings over {a, b} such that the number of b is the
double of the number of a. Let
 L1 = (anb2n) ,
which is non-regular, and
L2 = a*b*,
which is regular
 then L1 = L  L2
 By
closure properties of RLs:
if L is regular, then L1 is also regular
But since L1 not regular, so L can’t be regular

42


4.3: Pumping Lemma (21)

Common Mistakes
 Let L1 regular language, L3 and L4 are not regular.
 Is L regular?
 L = L1 ⊕ L3
0n1n = 0*1*  0n1n
0*1* = 0*1*  0n1n
L
= L3 ⊕ L4
0*1* = 0p1q  0q1p , where p  q
 L3
= L ⊕ L4
0n1n = 0*1*  0n1n

43



Class Discussion

Prove that L = {ww : w  (0|1)*} is non-regular

Is L = {(01)6n : n  0} regular?

Is L = {(ab)n : n  0} regular?
44




Summary

We use the Pumping Lemma to prove a language is
not regular
 Note, does not work for all non-regular
languages, though

Choosing a good string w is first key step

Choosing a good integer i is second key step

Must apply argument to all legal u, v, z
45

