Question

The number of operations executed by algorithms \(A\) and \(B\) is \(40 n^{2}\) and \(2 n^{3},\) respectively. Determine \(n_{0}\) such that \(A\) is better than \(B\) for \(n \geq n_{0}\)

Short answer

For \(n \geq 20\), algorithm A is better than algorithm B.

Step-by-step solution

Set Inequality

To find when algorithm A is better than algorithm B, set up the inequality: \(40n^2 < 2n^3\).

Simplify the Inequality

Divide both sides of the inequality by 2 to simplify: \(20n^2 < n^3\).

Rearrange the Inequality

Rearrange the inequality to isolate the term involving \(n\) on one side: \(n^3 - 20n^2 > 0\).

Factor the Polynomial

Factor the polynomial to find the roots: \(n^2(n - 20) > 0\).

Determine the Critical Points

Set each factor equal to zero to find the critical points: \(n^2 = 0\) (which gives \(n = 0\)), and \(n - 20 = 0\) (which gives \(n = 20\)).

Identify the Valid Range

Since \(n\) must be greater than 0 for algorithms to make sense, the meaningful root for our inequality is \(n = 20\). Therefore, for \(n \geq 20\), algorithm A is better than algorithm B.

Key concepts

algorithm efficiency

Algorithm efficiency measures how effectively an algorithm performs in terms of time and space usage. It helps us evaluate which algorithm will be faster or use fewer resources. The efficiency is often expressed as a function of input size, denoted as \(n\).

In the exercise, we compare two algorithms, \(A\) and \(B\), where their performance is given by \(40n^2\) and \(2n^3\) operations, respectively. The function \(40n^2\) means that the number of operations scales quadratically with input size, while \(2n^3\) scales cubically.

Quadratic functions generally grow more slowly than cubic functions. Hence, for large input sizes, an algorithm with a quadratic function will perform better (i.e., require fewer operations) than one with a cubic function.

complexity analysis

Complexity analysis is crucial for predicting the runtime and space requirements of an algorithm relative to its input size. Analyzing complexity helps identify inefficient algorithms and optimizes performance.

The problem involves comparing the time complexities of two algorithms, \(A\) and \(B\). Algorithm A has a complexity of \(O(n^2)\), and Algorithm B has a complexity of \(O(n^3)\). These Big-O notations indicate the upper-bound growth rates of the algorithms.

By setting up the inequality \(40n^2 < 2n^3\), we translated the problem into determining when Algorithm A surpasses Algorithm B in efficiency. As we simplify and solve the inequality, we find that for \(n \geq 20\), \(40n^2\) becomes less than \(2n^3\). Therefore, for large input sizes, Algorithm A is more efficient.

inequality solving

Solving inequalities is a fundamental skill in analyzing and comparing algorithms. It helps to determine the range of input sizes for which one algorithm outperforms another.

To compare algorithms \(A\) and \(B\), we start with the inequality \(40n^2 < 2n^3\). We simplify this by dividing both sides by 2, resulting in \(20n^2 < n^3\). Rearranging to \(n^3 - 20n^2 > 0\) facilitates factoring into \(n^2(n - 20) > 0\).

Finding the critical points, \(n = 0\) and \(n = 20\), allows us to identify intervals where the inequality holds true. Since \(n\) must be positive for practical use, the relevant solution is \(n \geq 20\). This defines the threshold beyond which Algorithm A is more efficient than Algorithm B.