Chapter 8.3, Problem 46E

To determine

Whether search of order $n{\log }_{2}n$ is more effective than the search of order $n^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ .

Answer to Problem 46E

Yes, the search with the order $n{\log }_{2}n$ is the most effective.

Explanation of Solution

Given information : The orders of the search of all the algorithms are $n{\log }_{2}n\text{ and }{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ .

An algorithm is effective in the long run if the number of searches is less as compared to other algorithms.

Compare the orders of search to find which of the three orders grows at a minimum rate.

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{n{\log }_{2}n}{n^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\underset{n\to \infty }{\lim }\frac{{\log }_2\left(n\right)+\frac{1}{\ln \left(2\right)}}{\frac{3}{2}{n}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ =\underset{n\to \infty }{\lim }\left(\frac{2{\log }_{2}n}{3\sqrt{n}}+\frac{2}{3\ln \left(2\right)\sqrt{n}}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{2\ln \left(2\right){\log }_{2}n+2}{3\ln \left(2\right)\sqrt{n}}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{2\ln \left(n\right)+2}{3\ln \left(2\right)\sqrt{n}}\right)\end{array}$

Solve further,

  $\begin{array}{c}\underset{n\to \infty }{\lim }\left(\frac{2\ln \left(n\right)+2}{3\ln \left(2\right)\sqrt{n}}\right)=\underset{n\to \infty }{\lim }\left(\frac{\frac{2}{n}}{3\ln \left(2\right)\times \frac{1}{2}{n}^{-1/2}}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{4}{3\ln \left(2\right)\sqrt{n}}\right)\\ =\frac{4}{3\ln \left(2\right)}\underset{n\to \infty }{\lim }\left(\frac{1}{\sqrt{n}}\right)\\ =0\end{array}$

Since the limit $\underset{x\to \infty }{\lim }\frac{n{\log }_{2}n}{n^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ is 0, the growth rate of $n{\log }_{2}n$ is less than that of $n^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ .

Consequently, the algorithm with the order of search $n{\log }_{2}n$ is more effective than that with the order $n^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ .