Operations on Regular Languages | Union | Intersection | Concatenation | Kleene Closure | Complement | Power of a String:
1. Union : If L1 and If L2 are two regular languages, their union L1 ∪ L2 will also be regular.
For example, L1 = {an | n ≥ 0} and L2 = {bn | n ≥ 0}
L3 = L1 ∪ L2 = {an ∪ bn | n ≥ 0} is also regular.
2. Intersection : If L1 and If L2 are two regular languages, their intersection L1 ∩ L2 will also be regular.
For example,
L1= {am bn | n ≥ 0 and m ≥ 0} and L2= {am bn ∪ bn am | n ≥ 0 and m ≥ 0}
L3 = L1 ∩ L2 = {am bn | n ≥ 0 and m ≥ 0} is also regular.
3. Concatenation : If L1 and If L2 are two regular languages, their concatenation L1.L2 will also be regular.
For example,
L1 = {an | n ≥ 0} and L2 = {bn | n ≥ 0}
L3 = L1.L2 = {am . bn | m ≥ 0 and n ≥ 0} is also regular.
4. Kleene Closure : If L1 is a regular language, its Kleene closure L1* will also be regular.
For example,
L1 = (a ∪ b)
L1* = (a ∪ b)*
5. Complement : If L(G) is regular language, its complement L’(G) will also be regular. Complement of a language can be found by subtracting strings which are in L(G) from all possible strings.
For example,
L(G) = {an | n > 3}
L’(G) = {an | n <= 3}
Note :
Two regular expressions are equivalent if languages generated by them are same. For example, (a+b*)* and (a+b)* generate same language. Every string which is generated by (a+b*)* is also generated by (a+b)* and vice versa.
Power of a String:
Consider ∑ = {0, 1}. The following are the power of an alphabet over input alphabet ∑.
(L)∑ = {null}: zero length string
(L)∑ = {0, 1}: set of 1 length string
(L)∑ = {01, 10, 11}: set of 2 length string
