Question

For each $\mathbf{n}$, exhibit two regular expressions,$\mathbf{R}\mathbf{and}\mathbf{S}$ , of length $\mathbf{poly}\mathbf{\left(}\mathbf{n}\mathbf{\right)}$, where$\mathbf{L}\mathbf{\left(}\mathbf{R}\mathbf{\right)}\mathbf{\ne }\mathbf{L}\mathbf{\left(}\mathbf{S}\mathbf{\right)}$, but where the first string on which they differ is exponentially long. In other words,$\mathbf{L}\mathbf{\left(}\mathbf{R}\mathbf{\right)}\mathbf{and}\mathbf{L}\mathbf{\left(}\mathbf{S}\mathbf{\right)}$ must be different yet agree on all strings of length up to${\mathbf{2}}^{\mathbf{nd}}$ for some constant $\mathbf{\epsilon }\mathbf{>}\mathbf{0}$.

Short answer

Every string of the length up to$2^{\mathrm{\epsilon n}}$ for the constant$\mathrm{\epsilon }>0$ , has been agreed for two regular expression$\mathrm{S}\mathrm{and}\mathrm{R}$ , where$\mathrm{L}\left(\mathrm{R}\right)\mathrm{and}\mathrm{L}\left(\mathrm{S}\right)$ must be different.

Step-by-step solution

To Regular the expression

The two regular expression$\mathrm{S}\mathrm{and}\mathrm{R}$ of the length$\mathrm{poly}\left(\mathrm{n}\right)$ has been exhibited for the$\mathrm{n}$ where$\mathrm{L}\left(\mathrm{R}\right)\ne \mathrm{L}\left(\mathrm{S}\right)$ in which they differ in exponentially long.

To Explain the polynomial length and constructed the expression

The polynomial length has been constructed in a way to recognize the $1^{\mathrm{k}}$for each $\mathrm{k}$and expression which are available of every first $\mathrm{n}$prime.

Every $\mathrm{n}-6$, the ${\mathrm{n}}^{\mathrm{th}}$ prime is smaller than$\mathrm{n}(\mathrm{lnn}+\mathrm{lnlnn})$ . As a result, the first $\mathrm{n}$ primes are added in $\mathrm{O}\left({\mathrm{n}}^2\mathrm{logn}\right)$ steps.

Because the first number that isn't a multiple of either of the first prime numbers has an order of $\mathrm{O}\left({\mathrm{n}}^2\mathrm{logn}\right)$, it can't grow exponentially with $\mathrm{n}$.