91Ó°ÊÓ

A language is regular if it can be expressed in terms of regular expression. A regular expression can also be described as a sequence of pattern that defines a string. Regular expressions are used to match character combinations in strings

$L\left(G\right)$is a set of strings of zeros and that either contain exactly 2$\#s$ and any number of zeros or contain exactly $1\#$and the number of 0s to the right of the$\#$ is twice the of zeros to the left.]

The given grammar does not follow the property of a regular language, so this is not a regular language.

$L\left(G\right)$is not regular.

Let prove this using closure properties. Here consider the following sequence of languages.

Now consider a homomorphism,

$h1:\{0,1,\#\}→\{0,\#\}$

such that,

$h1\left(0\right)=0,h1(\#)=\#,$ and $h1\left(1\right)=00.LetK1=h-11\left(L\left(G\right)\right).$

Let $K_2=K_1âˆ\copyright L(0*\#1*).$

Observe that the only strings in $L\left(G\right)$that have a single $\#$symbol are those belonging to $L_2$(defined in the previous part), and given the definition of homomorphism h1,

Here let’s conclude that $K_2=\left\{0i\#1i|iâ‰\yen 0\right\}.$

Consider the homomorphism $h2:\{0,1,\#\}→\{0,1\},whereh2\left(0\right)=0,h2\left(1\right)=1$

and then observe that $K3=h2\left(K2\right)=\left\{0^n{1}^n|nâ‰\yen 0\right\}$

Since $K1,K2,andK3$are obtained from$L\left(G\right)$ through regularity-preserving operations and $K3$is non-regular, we can conclude that$L\left(G\right)$ is not regular.

Hence,$L\left(G\right)$ is not regular language is proved.