## Example:

Q.  is not a regular language.

Solutions:

Initially, we always assume that it is regular language and the machine is DFA.

Let n is no. of state at DFA which accepts Language.

Now decompose it by xyz with two conditions.

Here,

Now we are going to check,

This time we check for  .

So, we can say,

Also | x | + | y | + | z | < | x | + 2| y | + | z | hence, | y | > 0 (true) or | y | ≥ 1 and | xy | ≤ n satisfy.

Now we can say,

| xy | ≤  n, | y | > 0 or | y | ≥ 1 then we can say | y | ≤ n because | xy | ≤  n.

Then we can say,

Now we can add (n + 1) on left-hand side of $\mathbf{|}{\mathbf{xy}}^{\mathbf{2}}\mathbf{z}\mathbf{|}$ then we can say,

Now looks at the given language  from language it is clear that every string ω  which belongs to the language is a length of ${\mathbf{i}}^{\mathbf{2}}$ (perfect square)

So if  then next string  here ${\mathbf{n}}^{\mathbf{2}}$ and  is perfect square.

But length of string ${\mathbf{xy}}^{\mathbf{2}}\mathbf{z}$  is between ${\mathbf{n}}^{\mathbf{2}}$ and   i.e. .

So, it is clear that $\mathbf{|}{\mathbf{xy}}^{\mathbf{2}}\mathbf{z}\mathbf{|}$ is not perfect square because between two perfect square no perfect square is possible like here ${\mathbf{n}}^{\mathbf{2}}$ and  are both perfect squares so,

Like     between 9 and 16 no perfect square is possible.

So,  ${\mathbf{xy}}^{\mathbf{2}}\mathbf{z}$ string violating pumping lemma, so,   violet pumping lemma and it is not Regular Language.