Question

Theoretical Foundation of Computer Science.

Is this a regular language: a set consisting of strings x such that x is of prime length or x is of odd length. Prove your answer.

Answer 1

This is a language L consists of 2 other languages L1 : A set of strings X such that is X is of Prime length or L2 : Set of strings X such that X is of odd length.

L = L1 ? L2

For L to be regular both L1 and L2 should be regular.

A language is regualr if we can define a FINITE AUTOMATA for it.

L1 is not regular as we can't define a FINTITE AUTOMATA for it.

we can prove this by using PUMPING LEMMA.

The pumping lemma says that for any regular language L there exists a constant p such that any word w in L with length at least p can be split into three substrings,
w = xyz, where the middle portion y must not be empty, such that the words xz, xyz, xyyz, xyyyz, … constructed by repeating y zero or more times are still in L.

L1 = { X | K is length of X and K is a prime number}

RULE :
i.e w = word or input belongs to L1 / |w| >=n
   w=xyz is in L1
   For all i ? 0, the string xyⁱz is also in L1

STEP 1 : Assume the given language is regular.

STEP2: As per pumping lemma , if the given language is regular, we can split the input w into 3 letters as xyz.

             w = xyz

             Length of xyz = K

             |xyz| = K and |y|>=1 {As middle portion y must not be empty}

            Now we will loop the middle portion y and check if it is belongs to the language of not.

           |xyⁱz| = K , here i should >=0 as mentioned earlier.

          So i value can also be K+1

        i.e |xy^K+1z| = |xyz|+|y^K|

                         = K+ K(|y|) { As we already know length of xyz is K and y is repeating K times}

             |xyⁱz| = K(1+|y|) {here lenght of y is not zero as per pumping lemma}

          we are represening the value in mulples of some numbers , that means it is not a prime number.

    As our state is not true, it is not REGULAR.

Even L2 is REGULAR as we can define a Finite automata but L1 is NOT REGULAR, L is NOT REGULAR.