Chapter 9.3, Problem 43E

To determine

To find that among $n{\log }_{2}n,n^{\frac{3}{2}},n{\left({\log }_{2}n\right)}^2$ which of the algorithm is likely the most efficient in the long run

Answer to Problem 43E

  $\left(n{\log }_{2}n\right)$ is more efficient in long run because it grows more slowly

Explanation of Solution

Given information:

The algorithm that grows more slowly is said to be more efficient so to find this each algorithm is calculated,

Now,

  $\begin{array}{l}n{\log }_{2}n\\ \underset{x\to \infty }{\lim }\frac{n}{{\log }_{2}n}\\ \underset{x\to \infty }{\lim }\frac{1}{\frac{1}{n{\log }_2}}=\infty \end{array}$

The function $n^{\frac{3}{2}}$ grows without any bound.

Therefore,

  $\underset{x\to \infty }{\lim }{n}^{\frac{3}{2}}=\infty$

And,

  $\begin{array}{l}n{\left({\log }_{2}n\right)}^2\\ \underset{x\to \infty }{\lim }\frac{n}{{\left({\log }_{2}n\right)}^2}\\ \underset{x\to \infty }{\lim }\frac{1}{\frac{1}{{\left(n{\log }_2\right)}^2}}\\ \underset{x\to \infty }{\lim }\frac{1}{\frac{1}{2n{\log }_2}}=\infty \end{array}$

Therefore, $\left(n{\log }_{2}n\right)$ is more efficient because it grows more slowly as $n\to \infty$