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.

For the given language, $A_{\frac{1}{3}-\frac{1}{3}}=\left\{xz\left|forsomey,\right|x\left|=\right|y\left|=\right|z\right|andxyz∈A\}$Let A be any language, define $A_{\frac{1}{3}-\frac{1}{3}}$be the subset of strings of AA whose middle third is removed.

The solution is came across makes the following claim,

which cannot be justify ,

Let $A=\{0*1*\}$then $A_{\frac{1}{3}-\frac{1}{3}}$$âˆ\copyright$$\{0*1*\}$$=\{0^n{1}^nâˆ\pounds nâ\copyright ¾0\}$

For example consider the string $w=0000\#1$, removing the middle third will yield $00\#1$and whose intersection with $\left\{0^{*}{1}^{*}\right\}$is∅which not of the form$\{0^n{1}^nâ\copyright ¾0\}$.

A string ss belongs to$A_{\frac{1}{3}-\frac{1}{3}}$iff you can form it from a string tt of the form $0^p\#1^q$by deleting the middle third t then here the string will be in$\{0*1*\}$iff the$\#$was in the middle third of t. The question is asking you to show that if that happens, then t contains an equal number of zeros and ones.

Hence, $A_{\frac{1}{3}-\frac{1}{3}}=\left\{xz\left|forsomey,\right|x\left|=\right|y\left|=\right|z\right|andxyz∈A\}$A is regular, then $A_{\frac{1}{3}-\frac{1}{3}}$is not necessarily regular is proved.