Question

Question: Let $\mathit{\Sigma }\mathbf{=}\left\{0,1\right\}$and let

$\mathit{D}\mathbf{=}\mathbf{\left\{}\mathbf{w}\mathbf{|}\mathbf{w}\mathbf{}\mathbf{contains}\mathbf{}\mathbf{ane}\mathbf{}\mathbf{qualnumber}\mathbf{}\mathbf{of}\mathbf{}\mathbf{occurrences}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{substrings}\mathbf{}\mathbf{01}\mathbf{}\mathbf{and}\mathbf{}\mathbf{10}\mathbf{\right\}}\mathbf{.}$

Thus$\mathbf{101}\mathbf{\in }\mathit{D}$ because 101 contains a single 01 and a single 10, but $\mathbf{1010}\mathbf{\notin }\mathit{D}$because 1010 contains two 10 s and one .01 Show that D is a regular language.

Short answer

Answer:

Regular language and its deterministic finite machine are shown below.

Step-by-step solution

Regular language

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.

Deterministic finite machine for regular language.

For the string which contains$\Sigma =\left\{0,1\right\}$ where$101\in D$because 101 contains a single 01 and a single 10 , but $1010\notin D$because 1010 contains two 10 s and one 01.

Hence, $D=\left\{w|wcontainsanequalnumberofoccurrencesofthesubstrings01and10\right\}$is a regular language and deterministic finite machine is possible.