Closure Properties of Regular
Languages
Union, Intersection, Difference,
Concatenation, Kleene Closure,
Reversal, Homomorphism, Inverse
Homomorphism
1
Review Closure Properties
Recall a closure property is a statement
that a certain operation on languages,
when applied to languages in a class
(e.g., the regular languages), produces
a result that is also in that class.
For regular languages, we can use any
of its representations to prove a closure
property.
2
Closure Under Union
If L and M are regular languages, so is
L  M.
Proof: Let L and M be the languages of
regular expressions R and S,
respectively.
Then R+S is a regular expression
whose language is L  M.
Illustrates use of RE representation
3
Closure Under Concatenation
and Kleene Closure
Same idea:
 RS is a regular expression whose language
is LM; therefor LM is regular.
 R* is a regular expression whose language
is L*; therefor L* is regular
4
Closure Under Intersection
If L and M are regular languages, then LM is
regular.
Proof: Let A and B be DFA’s whose languages
are L and M, respectively.
Construct C = p-DFA of A and B
 States of C are distinct pairs [q(A),r(B)]
 Start state is pair [start(A), start(B)]
 δ([q(A),r(B)],a) = [δA(q,a), δB(r,a)]
Make the accepting states of C be the pairs
consisting of accepting states of both A and B.
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Product DFA for Intersection
0
A
DFA(L)
1
P-DFA(L  M )
B
[A,C]
0, 1
1
1
0
[A,D]
1
0
DFA(M)
1
C
0
1
0
D
1
[B,C]
0
0
[B,D]
String w accepted by p-DFA iff
it is accepted by both DFA(L)
and DFA(M). P-DFA defines
(LM ), which is regular
6
Closure Under Difference
If L and M are regular languages, then so is
L – M = strings in L but not M.
Proof: Let A and B be DFA’s whose
languages are L and M, respectively.
Construct C = p-DFA of A and B.
Make the accepting states of C be the pairs
where A-state is accepting but B-state is not.
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Product DFA for Difference
0
A
DFA(L)
1
p-DFA(L-M)
B
[A,C]
0, 1
1
C
0
1
[A,D]
1
0
DFA(M)
1
1
0
0
1
[B,C]
0
0
[B,D]
D
DFA(L-M) is the empty language
in this case.
How do we know this?
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Closure Under Complementation
The complement of a language L (with
respect to an alphabet Σ such that Σ*
contains L) is Σ* – L.
Since Σ* is regular, the complement of
a regular language is regular because it
is the difference of regular languaes.
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Closure Under Reversal
Given language L, LR is the set of strings
whose reversal is in L.
Example: L = {0, 01, 100};
LR = {0, 10, 001}.
Proof: Let E be a regular expression for L.
We show how to reverse E, to provide a
regular expression ER for LR.
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Reversal of a Regular Expression
Basis: If E is a symbol a, ε, or
ER = E.
Induction:
∅,
then
 If E=F+G, then ER = FR + GR.
 If E=FG, then ER = GRFR
 If E=F*, then ER = (FR)*.
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Example: Reversal of a RE
Let E = 01* + 10*.
ER = (01* + 10*)R = (01*)R + (10*)R
= (1*)R0R + (0*)R1R
= (1R)*0 + (0R)*1
= 1*0 + 0*1.
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Homomorphisms
A homomorphism on an alphabet is a
function that assigns a string to every
symbol in that alphabet
 Example on S={0,1}
 Define h(0) = ab h(1) = ε
 Extend to strings by h(a1…an) = h(a1)…h(an)
therefore: h(01010) = ababab
h(L)={h(w)|w is in L}=homomorphism of L
 Language formed by applying h to every string
in L
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Closure Under Homomorphism
If L is a regular language, and h is a
homomorphism on its alphabet, then
h(L)= {h(w)|w is in L} is also a regular
language.
Proof: Let E be a regular expression for L.
Apply h to each symbol in E.
Language of resulting RE is h(L).
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Exercise: find the RE of h(L)
Let h(0) = ab; h(1) = ε.
Let L be the language of regular
expression 01* + 10*.
Then h(L) is the language of regular
expression abε* + ε(ab)*.
Note: use parentheses
to enforce the proper
grouping.
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Simply RE for h(L)
abε* + ε(ab)*
ε* = ε, so abε* = abε.
ε is the identity under concatenation.
abε*+ε(ab)*=abε+ε(ab)*=ab+(ab)*.
ab is contained in (ab)*
RE for h(L) is (ab)*
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Inverse Homomorphism of a string
h-1(w) is read “inverse homomorphism of w
w’=h-1(w), iff h(w’)=w
Testing an inverse homomorphism
candidate is easy.
Just apply homomorphism to the candidate
17
Inverse homomorphisms of a
language
Let h be a homomorphism defined on S
Let h(L) be a homomorphism of L
defined on S
Let h-1(L) denote the inverse
homomorphism of L defined by h(L)
h-1(L) ={w’| such that h(w’) is in L}
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Example: Inverse Homomorphism
Let h(0) = ab; h(1) = ε.
Let L = {abab, baba}
h-1(L) = {w defined on {0,1} such that
h(w) is either abab or baba}
No w such that h(w)=baba
For any w with two 0’s and any number
of 1’s h(w)=abab
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Closure of RLs under Inverse Homomorphism
Proof by construction
Start with a DFA = A for L
Construct the ih-DFA = B for h-1(L) with:
 The same set of states.
 The same start state.
 The same final states.
 Input alphabet = the symbols to which
homomorphism h applies
 Transition function δB(q, a) = δA(q, h(a))
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Example of ih-DFA Construction
DFA defined on {a,b}
ih-DFA defined on {0,1}
1
a
B
a
A
b
0
A
0
b
a
B
1
b
Since
h(1) = ε
C
δB(q, a) = δA(q, h(a))
h(0) = ab h(1) = ε
Since
h(0) = ab
C
1, 0
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CptS 317 Fall 2014
Assignment 9, Due 11-7-14
Exercises 4.2.1 (a), (c) and (e)
text p 147
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Quotient of a language L/a
L/a = set of all strings w such that wa is in L
a is a symbol in the alphabet of L
a is the last symbol in the test string wa
Example: if L={a,aab,baa} then L/a ={e,ba}
If L is regular, so is L/a
Proof by construction: q-DFA(L/a) same as
DFA(L) except that q is an accepting state of
q-DFA iff d(q,a) is an accepting state of DFA
illustrate
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Derivative of a language: a\L
a\L = set of all strings w such that aw is in L
a is the first symbol in the test string aw
Example: if L={a,aab,baa} then a\L={e,ab}
Called “derivative” because RE of L compared
to RE a\L is similar to derivative of algebraic
expression
Example: a\(R+S)=a\R+a\S remove a after
union is the same as union of strings with a
removed
Concatenation example on board
24
Exercise 4.2.5e
What is the implication of 0\L = nil?
Exercise 4.2.5f
Find a language such that 0\L = L
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Using Closure to prove non-regular
L1 is language in question
L2 is language known to be non-regular
O is an operation under which regular
languages are closed.
L2 = O L1
If L1 were regular, then L2 would be regular,
but it isn’t.
Therefore L1 is not regular
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Using Closure to prove non-regular
Example: 4.2.13 p149
Prove L1={0i1j|i>0,j>0,i not = j} not reg.
Assume L1 is regular
Then L2=0*1*-L1 is regular by closure
under difference
L2={0n1n|n>0} why?
L2={0n1n|n>0} is not regular
Therefor the assumption about L1 is false
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