Closure properties of Regular languages - GeeksforGeeks

15 Jan, 2020
Closure properties on regular languages are defined as certain operations on regular language which are guaranteed to produce regular language. Closure refers to some operation on a language, resulting in a new language that is of same “type” as originally operated on i.e., regular.
Regular languages are closed under following operations.
Consider L and M are regular languages:
Kleen Closure:RS is a regular expression whose language is L, M. R* is a regular expression whose language is L*.Positive closure:RS is a regular expression whose language is L, M.  is a regular expression whose language is .Complement:The complement of a language L (with respect to an alphabet  such that  contains L) is –L. Since  is surely regular, the complement of a regular language is always regular.Reverse Operator:Given language L,  is the set of strings whose reversal is in L.Example: L = {0, 01, 100}; ={0, 10, 001}.Proof: Let E be a regular expression for L. We show how to reverse E, to provide a regular expression  for .Complement:The complement of a language L (with respect to an alphabet  such that  contains L) is –L. Since  is surely regular, the complement of a regular language is always regular.Union: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 U M).Intersection:Let L and M be the languages of regular expressions R and S, respectively then it a regular expression whose language is L intersection M.proof: Let A and B be DFA’s whose languages are L and M, respectively. Construct C, the product automaton of A and B make the final states of C be the pairs consisting of final states of both A and B.Set Difference operator: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, the product automaton of A and B make the final states of C be the pairs, where A-state is final but B-state is not.Homomorphism:A homomorphism on an alphabet is a function that gives a string for each symbol in that alphabet. Example: h(0) = ab; h(1) = . Extend to strings by h(a1...an) =h(a1)...h(an). Example: h(01010) = ababab.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 R, E is h(L).Inverse Homomorphism : Let h be a homomorphism and L a language whose alphabet is the output language of h.  (L) = {w | h(w) is in L}.
Kleen Closure:RS is a regular expression whose language is L, M. R* is a regular expression whose language is L*.
Positive closure:RS is a regular expression whose language is L, M.  is a regular expression whose language is .
Complement:The complement of a language L (with respect to an alphabet  such that  contains L) is –L. Since  is surely regular, the complement of a regular language is always regular.
Reverse Operator:Given language L,  is the set of strings whose reversal is in L.Example: L = {0, 01, 100}; ={0, 10, 001}.Proof: Let E be a regular expression for L. We show how to reverse E, to provide a regular expression  for .
Complement:The complement of a language L (with respect to an alphabet  such that  contains L) is –L. Since  is surely regular, the complement of a regular language is always regular.
Union: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 U M).
Intersection:Let L and M be the languages of regular expressions R and S, respectively then it a regular expression whose language is L intersection M.proof: Let A and B be DFA’s whose languages are L and M, respectively. Construct C, the product automaton of A and B make the final states of C be the pairs consisting of final states of both A and B.
Set Difference operator: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, the product automaton of A and B make the final states of C be the pairs, where A-state is final but B-state is not.
Proof: Let A and B be DFA’s whose languages are L and M, respectively. Construct C, the product automaton of A and B make the final states of C be the pairs, where A-state is final but B-state is not.
Homomorphism:A homomorphism on an alphabet is a function that gives a string for each symbol in that alphabet. Example: h(0) = ab; h(1) = . Extend to strings by h(a1...an) =h(a1)...h(an). Example: h(01010) = ababab.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 R, E is h(L).
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 R, E is h(L).
Inverse Homomorphism : Let h be a homomorphism and L a language whose alphabet is the output language of h.  (L) = {w | h(w) is in L}.
Note: There are few more properties like symmetric difference operator, prefix operator, substitution which are closed under closure properties of regular language.
Decision Properties:Approximately all the properties are decidable in case of finite automaton.
(i) Emptiness 
(ii) Non-emptiness 
(iii) Finiteness 
(iv) Infiniteness 
(v) Membership 
(vi) Equality  
These are explained as following below.
(i) Emptiness and Non-emptiness:
Step-1: select the state that cannot be reached from the initial states & delete them (remove unreachable states).
Step 2: if the resulting machine contains at least one final states, so then the finite automata accepts the non-empty language.
Step 3: if the resulting machine is free from final state, then finite automata accepts empty language.(ii) Finiteness and Infiniteness:Step-1: select the state that cannot be reached from the initial state & delete them (remove unreachable states).Step-2: select the state from which we cannot reach the final state & delete them (remove dead states).Step-3: if the resulting machine contains loops or cycles then the finite automata accepts infinite language.Step-4: if the resulting machine do not contain loops or cycles then the finite automata accepts infinite language.(iii) Membership:Membership is a property to verify an arbitrary string is accepted by a finite automaton or not i.e. it is a member of the language or not.Let M is a finite automata that accepts some strings over an alphabet, and let ‘w’ be any string defined over the alphabet, if there exist a transition path in M, which starts at initial state & ends in anyone of the final state, then string ‘w’ is a member of M, otherwise ‘w’ is not a member of M.(iv) Equality:Two finite state automata M1 & M2 is said to be equal if and only if, they accept the same language. Minimise the finite state automata and the minimal DFA will be unique.My Personal Notes
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(ii) Finiteness and Infiniteness:
Step-1: select the state that cannot be reached from the initial state & delete them (remove unreachable states).
Step-2: select the state from which we cannot reach the final state & delete them (remove dead states).
Step-3: if the resulting machine contains loops or cycles then the finite automata accepts infinite language.
Step-4: if the resulting machine do not contain loops or cycles then the finite automata accepts infinite language.
(iii) Membership:Membership is a property to verify an arbitrary string is accepted by a finite automaton or not i.e. it is a member of the language or not.
Let M is a finite automata that accepts some strings over an alphabet, and let ‘w’ be any string defined over the alphabet, if there exist a transition path in M, which starts at initial state & ends in anyone of the final state, then string ‘w’ is a member of M, otherwise ‘w’ is not a member of M.
(iv) Equality:Two finite state automata M1 & M2 is said to be equal if and only if, they accept the same language. Minimise the finite state automata and the minimal DFA will be unique.
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