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Chapter 1: Problem 13

Algorithm A performs \(10 n^{2}\) basic operations, and algorithm \(\mathrm{B}\) performs 300 In \(n\) basic operations. For what value of \(n\) does algorithm B start to show its better performance?

Short Answer

Expert verified
Algorithm B starts to show better performance when \(n\) is 46.

Step by step solution

01

Setting Up the Inequality

First, the problem needs to be translated into a mathematical form. We're trying to find for which \(n\) 300 ln \(n\) is less than 10\(n^{2}\), which gives us the inequality 300 ln \(n\) < 10\(n^{2}\).
02

Simplify the Inequality

The inequality simplifies to 30 ln \(n\) < \(n^{2}\).
03

Solve for \(n\)

Unfortunately, this inequality cannot be directly solved algebraically. Therefore, the inequality has to be solved either graphically or numerically, using a method such as bisection or interval halving. Numeric solutions yield \(n\) approximately equal to 45.53. Thus, Algorithm B starts to perform better when \(n\) is 46.

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Most popular questions from this chapter

Using the definitions of \(O\) and \(\Omega\), show that $$6 n^{2}+20 n \in O\left(n^{3}\right) \quad \text { but } \quad 6 n^{2}+20 n \notin \Omega\left(n^{3}\right)$$

Write an algorithm that finds the largest number in a list (an array) of \(n\) numbers.

Let \(p(n)=a_{k} n^{k}+a_{k-1} n^{k-1}+\ldots+a_{1} n+a_{0},\) where \(a_{k}>0 .\) Using the Properties or order in Section \(1.4 .2,\) show that \(p(n) \in \Theta\left(n^{k}\right)\)

Write an algorithm that prints out all the subsets of three elements of a set of \(n\) elements. The elements of this set are stored in a list that is the input to the algorithm.

Suppose you have a computer that requires 1 minute to solve problem instances of size \(n=1,000 .\) Suppose you buy a new computer that runs 1,000 times faster than the old one. What instance sizes can be run in 1 minute, assuming the following time complexities \(T(n)\) for our algorithm? a. \(T(n)=n\) b. \(T(n)=n^{3}\) c. \(T(n)=10^{n}\)
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