/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Agorithms A and B porfom the sam... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Agorithms A and B porfom the same tardk. Cn inpur of s coe n, sigorithm A cwocutes 0.003 \(m^{2}\) instructions, and algorithm B expcutes 243 n inetructions. Find the approwimate value of \(n\) above ahich algorithm \(B\) is mono efficient. Prou may use a calculator or spreadsheet.

Short Answer

Expert verified
Algorithm B is more efficient for \(n > 81000\).

Step by step solution

01

Understand the Problem

We are given two algorithms, A and B. We need to find the value of \(n\) above which algorithm B is more efficient than algorithm A. This means we need to solve the inequality: Algorithm B's execution count should be less than Algorithm A's execution count.
02

Set up the Inequality

Algorithm A executes \(0.003 \cdot n^2\) instructions. Algorithm B executes \(243 \cdot n\) instructions. We want to find the smallest value of \(n\) such that \(243n < 0.003n^2\).
03

Simplify the Inequality

Rearrange the inequality \(243n < 0.003n^2\) to \(0.003n^2 - 243n > 0\).
04

Solve Quadratic Inequality

Factor the quadratic inequality. Start by dividing everything by 0.003: \(n^2 - 81000n > 0\). Simplify to find \(n(n - 81000) > 0\).
05

Identify Critical Points

The critical points of the inequality \(n(n - 81000) > 0\) are \(n = 0\) and \(n = 81000\). Test intervals around these points to find where the inequality is satisfied.
06

Determine Solution Interval

For \(n(n - 81000) > 0\), test the intervals \((0, 81000)\) and \((81000, \infty)\). At \(n = 81000\), expressions change sign, showing \(n > 81000\) satisfies the inequality.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Inequality
A quadratic inequality involves expressions where a quadratic term, typically denoted as \( ax^2 + bx + c \), is compared to a value, to determine an interval on the number line. In our problem, we worked with an inequality resulting from comparing the computational efficiency of two algorithms. We had to solve \( 0.003n^2 - 243n > 0 \). Simplifying this inequality helps us find the critical points by factoring or using the quadratic formula.
Quadratic inequalities differ from quadratic equations. Inequalities require us to find solution sets or intervals. Critical points (or roots) are often located where the expression changes signs. These situations may demand testing the inequality on different intervals between and beyond these roots.
To solve, first divide and simplify the inequality. For example, our inequality turned into \( n(n - 81000) > 0 \), marking \( n = 0 \) and \( n = 81000 \) as the critical points. Testing the solution in intervals, we identified \( n > 81000 \) as where algorithm B becomes more efficient.
Algorithm Comparison
Algorithm comparison is a process of evaluating different algorithms against one another to determine which performs more efficiently under given conditions. This is essential in computer science for optimizing operations. In our example, we needed to compare algorithm A and B to find out when algorithm B becomes more efficient based on their instruction count.
Algorithm A's computational requirement grows quadratically, represented by \(0.003n^2\), which means its performance will decline more significantly as \(n\) increases. In contrast, algorithm B's linear growth \(243n\) remains steadier on increases of \(n\).
The tipping point where B overtakes A in efficiency lies beyond a particular \(n\). This point, found through inequality solving, demonstrated that algorithm B becomes more efficient when \( n > 81000 \). Through comparison, you understand what makes an algorithm faster or slower depending on problem size \(n\).
Computational Complexity
Computational complexity offers a lens into the efficiency of algorithms, describing how the number of resources required for an algorithm scales with the input size \( n \). In solving our problem, computational complexity helped us understand the efficiency variance between two algorithms.
With algorithm A growing at \( O(n^2) \) and algorithm B at \( O(n) \), we see they have different efficiencies. The potential for decreased efficiency due to inputs increasing illustrates why understanding complexity is pivotal when evaluating algorithms. Where one might expect B's linear complexity to dominate A's quadratic complexity over large inputs, our task precisely verifies that for \( n > 81000 \), algorithm B indeed requires fewer computational resources than algorithm A.
The analysis aids decision-making on which algorithm to implement, depending on acceptable input sizes. This complexity-based forethought enables informed choices in designing or selecting algorithms tailored to specific environments or constraints.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A tennis tournarnsnt has 342 players. A singlo match imolves 2 players. The winner of a match will play the winner of a match in the next round, wheress losers are eliminated from the toumament. The 2 players who have won all previous rounds play in the final game, and the wirner wins the tournament. What is the total number of matches needed to determine the winner? a. Here is one algorithm to answer this question. Compute \(342 / 2=171\) to get the number of pairs (matches) in the first round, which results in 171 winners to go on to the second round. Compute \(171 / 2=85\) with 1 left over, which results in 85 matches in the second round and 85 winners, plus the 1 left over, to go on to the third round. For the third round compute \(86 / 2=43,50\) the third round has 43 matches, and so on. The total number of matches is \(171+85+43+\ldots\) Finish this process to find the total number of matches. b. Here is another algorithm to solve this problem, Each match results in exactly one loser, so there must be the same number of matches as losers in the tournament. Compute the total number of losers in the entire tournament. (Hint: This isf't really a computation; it is a one-sentence argument.) c. What is your opinion on the relative clarity, elegance, and efficiency of the two algorithms?

Perform a selection sort on the list 7, 4, 2, 9, 6. Show the list after each exchange that has an effect on the list orclering.

Consider the following sorted list of names. Arturo, Elsa, JoAnn, John, José, Lee, Snyder, Tracy a. Use binary search to decide whether Elsa is in this list. What names will be compared with Elsa? b. Use binary search to decide whether Tracy is in this list. What names will be compared with Tracy? c. Use binary search to decide whether Emile is in this list. What names will be compared with Emile?

In the Flipping Pancakes box, the original algorithm given requires at most \(2 n-3\) flips in the worst case. The claim is made that the new algorithm, which requires at most \(15 n+5] / 3\) flipa, is a better algorithm. How many pancalces do you need to hawe betore the second algorithm is indeed faster? Use a calculator or spreadsheet.]

Show the steps in merging \(A\) and \(B\) into \(C\) where $$ A=8,12,19,34 \quad B=3,5,15,21 $$

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.