1. Closure Properties of Regular Languages
Union, Intersection, Difference, Concatenation, Kleene Closure, Reversal, Homomorphism, Inverse Homomorphism

2. 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.

3. Closure Under Union If L and M are regular languages, so is L  M.
Proof: Let R and S be the REs that define L and M.
Then R+S is a regular expression whose regular language is L  M.
Therefore, RLs are closed under union

4. Closure Under Concatenation and Kleene Closure
Same idea:
RS is a regular expression whose language is LM; therefore LM is regular.
R* is a regular expression whose language is L*; therefore L* is regular

5. 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.

6. Product DFA for IntersectionB 1
0, 1
P-DFA(L  M ) L
[A,C]
[A,D] 1 1 1 1 C D
[B,C]
[B,D] 1 M
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

7. Closure Under Difference
L– M = strings in L but not M. If L and M are regular languages, then so is L-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.
p-DFA defines the RL of L-M.

8. Product DFA for DifferenceB 1
0, 1
p-DFA(L-M) L
[A,C]
[A,D] 1 1 1 C D 1
[B,C]
[B,D] 1 M
p-DFA(L-M) is the empty language
in this case, why?

9. 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.

10. 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.

11. Reversal of a Regular Expression
Basis: If E = a, ε, or ∅, then ER = E.
Induction:
If E=F+G, then ER = FR + GR.
If E=FG, then ER = GRFR
If E=F*, then ER = (FR)*.
Example find ER of 1*0 + 0*1. Show all steps

12. Example: Reversal of a RE
Let E = 01* + 10*.
ER = (01* + 10*)R
= (01*)R + (10*)R (union rule)
= (1*)R0R + (0*)R1R (concat. rule)
= (1R)*0 + (0R)*1 (closure rule)
= 1*0 + 0*1.

13. 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) example: h(01010) = ababab
h(L)={h(w)|w is in L}=homomorphism of L
Language formed by applying h to every string in L

14. 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).

15. Exercise: find the RE of h(L) and simplify
h(0) = ab; h(1) = ε.
L = the language of RE = 01* + 10*
Find the RE of h(L) and simplify

16. Simplify the RE for h(L)
abε* + ε(ab)*
ε* = ε
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)*

17. Inverse Homomorphism of a string
h-1(w) is read “inverse homomorphism of w
w’=h-1(w), iff h(w’)=w
Testing a string for inverse homomorphism is easy.
Just apply homomorphism to the string

18. 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}

19. Example: Inverse Homomorphism
Let h(0) = ab; h(1) = ε.
Let L = {abab, baba}
h-1(L) = {all w defined on {0,1} such that h(w) is either abab or baba}
No w such that h(w)=baba
h-1(L) = any w with two 0’s and any number of 1’s because h(w)=abab

20. Closure of RLs under Inverse Homomorphism Proof by construction
Start with a DFA(L) = A
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))

21. Example of ih-DFA Construction
ih-DFA defined on {0,1}
DFA defined on {a,b} 1
Since
h(1) = ε a B C B A
, 0
Since
h(0) = ab a A b b b C a
δB(q, a) = δA(q, h(a))
h(0) = ab h(1) = ε

22. CptS 317 Fall 2016
Assignment 12, Due
Exercises (a), (c) and (e) text p 147

23. 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

24. 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

25. Exercise 4.2.5e
What is the implication of 0\L = nil?
Exercise 4.2.5f
Example of L such that 0\L = L

26. 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

27. Using Closure to prove non-regular
Example: 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