Question

Let $\mathit{\Sigma }\mathbf{2}$be the same as in Problem 1.33. Consider each row to be a binary number and let $\mathit{D}\mathbf{=}\mathbf{\{}\mathit{w}\mathbf{\in }\mathit{\Sigma }\mathbf{*}\mathbf{2}\mathbf{|}$the top row of w is a larger number than is the bottom row}. For example, $\mathbf{00101100}\mathbf{\in }\mathit{D}$, but $\mathbf{000111006}\mathbf{\in }\mathit{D}$. How that D is regular.

Short answer

Dis a regular language.

Step-by-step solution

Considering each has binary number

The Given language is $D=\{w\in \sum _2^{*}|$

The top of w is the larger number than is the bottom row}

Over the alphabet is

The Language for given expression

Here each row has binary number.

Explanation

Let M be theDFA, over the input alphabet .

The state transition diagram ofM is as follows:

Prove that D is a regular language.

A language is said to be regular if it is recognized by a DFA.

Let take string form language,

Initial state of the aboveDFA is ‘equal’

Parse string

Here ‘large’ is final state , the string is accepted by the DFA.

Thus, language of given D is accepted by the given.

DFA to recognize the language D.

Therefore, D is a regular language.