Chapter 14, Problem 4RE

Explanation of Solution

Given:

The actual number of visit in the selection sort algorithm is :

$\text{T}\left(\text{n}\right)\text{=}\frac{\text{1}}{\text{2}}{\text{n}}^{\text{2}}\text{+}\frac{\text{5}}{\text{2}}\text{n-3}$

The order of growth for the above function is $\text{O}\left({\text{n}}^{\text{2}}\right)$.

Computation of the actual ratio:

$\text{T}\left(\text{n}\right)\text{=}\frac{\text{1}}{\text{2}}{\text{n}}^{\text{2}}\text{+}\frac{\text{5}}{\text{2}}\text{n-3}$

Substituting the value of n in the function provided.

$\begin{array}{l}\frac{\text{T}\left(\text{2000}\right)}{\text{T}\left(\text{1000}\right)}\\ =\frac{\frac{1}{2}{\left(2000\right)}^2+\frac{5}{2}\left(2000\right)-3}{\frac{1}{2}{\left(1000\right)}^2+\frac{5}{2}\left(1000\right)-3}\\ =\frac{2004997}{502497}\\ =3.99\end{array}$

In the below function the number of visits are present in the order $\text{O}\left({\text{n}}^{\text{2}}\right)$

$\text{f}\left(\text{n}\right){\text{=n}}^{\text{2}}$

$\begin{array}{l}\frac{f\left(\text{2000}\right)}{f\left(\text{1000}\right)}\\ =\frac{\frac{1}{2}{\left(2000\right)}^2}{\frac{1}{2}{\left(1000\right)}^2}\\ =4\end{array}$

$\text{T}\left(\text{n}\right)\text{=}\frac{\text{1}}{\text{2}}{\text{n}}^{\text{2}}\text{+}\frac{\text{5}}{\text{2}}\text{n-3}$

Substituting the value of n in the function provided.

$\begin{array}{l}\frac{\text{T}\left(\text{4000}\right)}{\text{T}\left(\text{1000}\right)}\\ =\frac{\frac{1}{2}{\left(4000\right)}^2+\frac{5}{2}\left(4000\right)-3}{\frac{1}{2}{\left(1000\right)}^2+\frac{5}{2}\left(1000\right)-3}\\ =\frac{8009997}{502497}\\ =15\end{array}$