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

2. 2 Review Closure Properties uRecall 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. uFor regular languages, we can use any of its representations to prove a closure property.

3. 3 Closure Under Union uIf L and M are regular languages, so is L  M. uProof: Let L and M be the languages of regular expressions R and S, respectively. uThen R+S is a regular expression whose language is L  M. uIllustrates use of RE representation

4. 4 Closure Under Concatenation and Kleene Closure uSame idea: wRS is a regular expression whose language is LM; therefor LM is regular. wR* is a regular expression whose language is L*; therefor L* is regular

5. 5 Closure Under Intersection uIf L and M are regular languages, then L  M is regular. uProof: Let A and B be DFA’s whose languages are L and M, respectively. uConstruct C = p-DFA of A and B wStates of C are distinct pairs [q(A),r(B)]  Start state is pair [start(A), start(B)]  δ ([q(A),r(B)],a) = [ δ A (q,a), δ B (r,a)] uMake the accepting states of C be the pairs consisting of accepting states of both A and B.

6. 6 Product DFA for Intersection AB 0 1 0, 1 1 CD 1 0 0 [A,C][A,D] 0 [B,C] 1 0 1 0 1 [B,D] 0 1 DFA(L) DFA(M) P-DFA(L  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. 7 Closure Under Difference uIf L and M are regular languages, then so is L – M = strings in L but not M. uProof: Let A and B be DFA’s whose languages are L and M, respectively. uConstruct C = p-DFA of A and B. uMake the accepting states of C be the pairs where A-state is accepting but B-state is not.

8. 8 Product DFA for Difference AB 0 1 0, 1 CD 1 1 0 0 [A,C][A,D] 0 [B,C] 1 0 1 0 1 [B,D] 0 1 DFA(L-M) is the empty language in this case. How do we know this? DFA(L) DFA(M) p-DFA(L-M)

9. 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. 10 Closure Under Reversal uGiven language L, L R is the set of strings whose reversal is in L. uExample: L = {0, 01, 100}; L R = {0, 10, 001}. uProof: Let E be a regular expression for L. uWe show how to reverse E, to provide a regular expression E R for L R.

11. 11 Reversal of a Regular Expression  Basis: If E is a symbol a, ε, or ∅, then E R = E. uInduction: wIf E=F+G, then E R = F R + G R. wIf E=FG, then E R = G R F R wIf E=F*, then E R = (F R )*.

12. 12 Example: Reversal of a RE Let E = 01* + 10*. E R = (01* + 10*) R = (01*) R + (10*) R = (1*) R 0 R + (0*) R 1 R = (1 R )*0 + (0 R )*1 = 1*0 + 0*1.

13. 13 Homomorphisms uA homomorphism on an alphabet is a function that assigns a string to every symbol in that alphabet  Example on  ={0,1}  Define h(0) = ab h(1) = ε wExtend to strings by h(a 1 …a n ) = h(a 1 )…h(a n ) therefore: h(01010) = ababab uh(L)={h(w)|w is in L}=homomorphism of L wLanguage formed by applying h to every string in L

14. 14 Closure Under Homomorphism uIf 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. uProof: Let E be a regular expression for L. uApply h to each symbol in E. uLanguage of resulting RE is h(L).

15. 15  Let h(0) = ab; h(1) = ε. uLet L be the language of regular expression 01* + 10*.  Then h(L) is the language of regular expression ab ε * + ε (ab)*. Note: use parentheses to enforce the proper grouping. Exercise: find the RE of h(L)

16. 16 Simply RE for h(L)  ab ε * + ε (ab)*  ε * = ε, so ab ε * = ab ε.  ε is the identity under concatenation.  ab ε *+ ε (ab)*=ab ε + ε (ab)*=ab+(ab)*. uab is contained in (ab)* uRE for h(L) is (ab)*

17. 17 Inverse Homomorphism of a string uh -1 (w) is read “inverse homomorphism of w uw’=h -1 (w), iff h(w’)=w uTesting an inverse homomorphism candidate is easy. uJust apply homomorphism to the candidate

18. 18 Inverse homomorphisms of a language  Let h be a homomorphism defined on   Let h(L) be a homomorphism of L defined on   Let h -1 (L) denote the inverse homomorphism of L defined by h(L) uh -1 (L) ={w’| such that h(w’) is in L}

19. 19 Example: Inverse Homomorphism  Let h(0) = ab; h(1) = ε. uLet L = {abab, baba} uh -1 (L) = {w defined on {0,1} such that h(w) is either abab or baba} uNo w such that h(w)=baba uFor any w with two 0’s and any number of 1’s h(w)=abab

20. 20 Closure of RLs under Inverse Homomorphism Proof by construction uStart with a DFA = A for L uConstruct the ih-DFA = B for h -1 (L) with: wThe same set of states. wThe same start state. wThe same final states. wInput alphabet = the symbols to which homomorphism h applies  Transition function δ B (q, a) = δ A (q, h(a))

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

22. 22 CptS 317 Fall 2014 Assignment 9, Due 11-7-14 Exercises 4.2.1 (a), (c) and (e) text p 147

23. 23 Quotient of a language L/a uL/a = set of all strings w such that wa is in L ua is a symbol in the alphabet of L ua is the last symbol in the test string wa  Example: if L={a,aab,baa} then L/a ={ ,ba} uIf 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  (q,a) is an accepting state of DFA uillustrate

24. 24 Derivative of a language: a\L ua\L = set of all strings w such that aw is in L ua is the first symbol in the test string aw  Example: if L={a,aab,baa} then a\L={ ,ab} uCalled “derivative” because RE of L compared to RE a\L is similar to derivative of algebraic expression uExample: a\(R+S)=a\R+a\S remove a after union is the same as union of strings with a removed uConcatenation example on board

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

26. 26 Using Closure to prove non-regular uL 1 is language in question uL 2 is language known to be non-regular uO is an operation under which regular languages are closed. uL 2 = O L 1 uIf L 1 were regular, then L 2 would be regular, but it isn’t. uTherefore L 1 is not regular

27. 27 Using Closure to prove non-regular uExample: 4.2.13 p149 uProve L1={0 i 1 j |i>0,j>0,i not = j} not reg. uAssume L1 is regular uThen L2=0*1*-L1 is regular by closure under difference uL2={0 n 1 n |n>0} why? uL2={0 n 1 n |n>0} is not regular uTherefor the assumption about L1 is false

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