Question

Give regular expressions with exponentiation that generate the following languages over the alphabet $\mathbf{\{}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{\}}$.

a. All strings of length 500

b. All strings of length 500 or less

c. All strings of length 500 or more

d. All strings of length different than 500

e. All strings that contain exactly 500 1s

f. All strings that contain at least 500 1s

g. All strings that contain at most 500 1s

h. All strings of length 500 or more that contain a 0 in the 500th position

i. All strings that contain two 0s that have at least 500 symbols between them

Short answer

Above problem is solved using the regular expressions. It is basically the combination of the string of the power length which is k and the power is denoted by the symbol $\sum$i.e., power of an alphabet is denoted by ${\sum }_{}^{\mathrm{k}}$.

a. Regular Expression with exponentiation: ${\sum }_{}^{500}$

b.Regular Expression with exponentiation:$\mathbf{(}\sum \mathbf{U\epsilon }{\mathbf{)}}^{500}$

c.Regular Expression with exponentiation:${\sum }^{500}{\sum }^{*}$

d. Regular Expression with exponentiation: $\mathbf{(}\sum \mathbf{U\epsilon }{\mathbf{)}}^{499}\mathbf{U}{\sum }^{501}{\sum }^{*}$

e. Regular Expression with exponentiation: ${\mathbf{\left(}\mathbf{0}\mathbf{\right)}}^n{\mathbf{\left(}\mathbf{1}\mathbf{\right)}}^{500}$ where, $\mathrm{n}\ge 0$

f. Regular Expression with exponentiation: ${\mathbf{0}}^n{\mathbf{1}}^{500}{\mathbf{1}}^{*}$, where $\mathrm{n}\ge 0$

g. Regular Expression with exponentiation: where,${\mathbf{0}}^n{\mathbf{\left(}\mathbf{1}\mathbf{U\epsilon }\mathbf{\right)}}^{500}$where, $\mathrm{n}\ge 0$

h. Regular Expression with exponentiation:$\mathbf{(}\sum {\mathbf{)}}^{499}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{(}\sum {\mathbf{)}}^{*}$

i.Regular Expression with exponentiation:$\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{(}\sum {\mathbf{)}}^{500}\mathbf{(}\sum {\mathbf{)}}^{*}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$

Step-by-step solution

Using Regular Expressions 

The searching is done very easily using the Expression called Regular Expression. It consists of normal or the combination of the special character. Regular Expression is used for matching the combination of character in the string.

a. Given:

Every string is of length 500

Regular Expression with exponentiation: ${\sum }_{}^{500}$

• So, from above it is clear that the value of k is 500 i.e. $\mathrm{k}=500$,
• Hence, it represents the combination of the 500 strings with exponentiation.

With the help of the Regular Expression solving the problems 

b. Given :

All strings of length 500 or less

Regular Expression with exponentiation:$\mathbf{(}\sum \mathbf{U\epsilon }{\mathbf{)}}^{500}$

• In the above expression, $\epsilon$is denoted by empty string and the union of $\epsilon$and exponentiation will be less than or equal to 500.
• If there is no empty string then the combination will be equal to 500 and if there is an empty string then the combination will be less than 500 strings.

c. Given:

All strings of length 500 or more

Regular Expression with exponentiation:${\sum }^{500}{\sum }^{*}$

Where,${\sum }^{*}$denotes all the strings which is more than 500 and the string of length 500 is denoted by ${\sum }^{500}$

The multiplication of both the above exponentiation will result the string of the length 500 or more than 500.

d. Given:

Every string should have length different than 500

Regular Expression with exponentiation:$\mathbf{(}\sum \mathbf{U\epsilon }{\mathbf{)}}^{499}\mathbf{U}{\sum }^{501}{\sum }^{*}$

Where, ${(\sum \cup \mathrm{\epsilon })}^{499}$denotes all the strings which is 499 or less than 499 and$\mathrm{\epsilon }$denotes an empty string.

${\sum }^{501}{\sum }^{*}$denotes all the strings which is equal to 501 and more than 501.

The Union of both the above expression gives the required answer i.e.,

All the strings of length different than 500.

e. Given:

All strings that contain exactly 500 1s

Regular Expression with exponentiation:${\mathbf{\left(}\mathbf{0}\mathbf{\right)}}^n{\mathbf{\left(}\mathbf{1}\mathbf{\right)}}^{500}$ where,$\mathbf{n}\mathbf{\ge }\mathbf{0}$

Here,${\text{(}0\text{)}}^{\mathrm{n}}{\text{(}1\text{)}}^{500}$ denotes all the strings which contains exactly 500 1’s

Continuing in solving the problem using the Regular Expression

f. Given:

All strings that contain at least 500 1s

Regular Expression with exponentiation: ${\mathbf{0}}^n{\mathbf{1}}^{500}{\mathbf{1}}^{*}$, where$\mathrm{n}\ge 0$

Here, all the string which contains 500 1’s or more than 500 1’s and number of 0’s in the string is greater than or equal to 0 is denoted by$0^n{1}^{500}{1}^{*}$.

g. Given:

All strings that contain at most 500 1s

Regular Expression with exponentiation:${\mathbf{0}}^n{\mathbf{\left(}\mathbf{1}\mathbf{U\epsilon }\mathbf{\right)}}^{500}$ where,$\mathbf{n}\mathbf{\ge }\mathbf{0}$

Here, all the string which contains less than or equal to 500 1’s and number of 0’s in the string is greater than or equal to 0 are denoted by$0^{\mathrm{n}}{\text{(}1\cup \mathrm{\epsilon }\text{)}}^{500}$ .

The combination will be equal to 500 only if there is no empty string.

h. Given :

All strings of length 500 or more that contain a 0 in the 500th position

Regular Expression with exponentiation:$\mathbf{(}\sum {\mathbf{)}}^{499}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{(}\sum {\mathbf{)}}^{*}$

Here,${\sum }^{*}$ denotes all the strings which is more than 500 and the string of the length 499 is denoted by ${\sum }^{499}$

i. Given :

All strings that contain two 0s that have at least 500 symbols between them

Regular Expression with exponentiation:$\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{(}\sum {\mathbf{)}}^{500}\mathbf{(}\sum {\mathbf{)}}^{*}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$

Here, ${\sum }^{*}$denotes all the strings which is more than 500 and there is minimum 500 1’s between two 0’s.