Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Fill in the blanks in each of the following statements: a. A selection sort application would take approximately ________ times as long to run on a 128-element vector as on a 32-element vector. b. The efficiency of merge sort is ________.

Short Answer

Expert verified
a. 16 times; b. \( O(n \, \log n) \).

Step by step solution

01

Understanding Selection Sort Complexity

Selection sort is a simple sorting algorithm with a time complexity of \( O(n^2) \), where \( n \) is the number of elements in the array or vector. This means that the running time increases quadratically with the number of elements.
02

Determine the Factor of Increase

If a 32-element vector requires \( k \) time to sort, then a 128-element vector would require \( k \cdot (128/32)^2 = k \cdot 16 \) times the time due to the quadratic nature of selection sort. Therefore, a selection sort would take approximately 16 times as long to run on a 128-element vector as on a 32-element vector.
03

Understanding Merge Sort Efficiency

Merge sort is a more efficient sorting algorithm with a time complexity of \( O(n \, \log n) \). This means the algorithm has a logarithmic factor in its growth rate, making it more efficient for larger datasets compared to quadratically scaling algorithms like selection sort.

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 Vaia!

Key Concepts

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

Selection Sort
Selection sort is a straightforward sorting algorithm that is often used for teaching and learning due to its simplicity. Yet, its efficiency is not optimal for large datasets. Here’s why:Selection sort operates by dividing the list or array into two parts: a sorted subsection and an unsorted subsection. Initially, the sorted subsection is empty:
  • The algorithm repeatedly selects the smallest (or largest, depending on order) element from the unsorted portion and swaps it with the first unsorted element.
  • This process is repeated, moving gradually through the array until all elements are sorted.
The time complexity of this sorting technique is quadratic, specifically, it is represented by the big O notation as \( O(n^2) \), where \( n \) is the number of items being sorted. For a 32-element vector, the time taken is proportional to \( 32^2 \). However, if you increase the vector to 128 elements, the time needed becomes \( 128^2 \), which is 16 times longer due to the nature of quadratic growth.
Merge Sort
Unlike selection sort, merge sort is known for its efficiency, especially with larger datasets. It is a divide-and-conquer algorithm that:
  • Splits the array into two halves.
  • Recursively sorts each half.
  • Merges the sorted halves back together into a fully sorted array.
The power of merge sort lies in its time complexity, which is \( O(n \log n) \). This complexity arises because the list is divided in half at each step (the \( \log n \) part) and each of the \( n \) elements needs to be considered in the merging process. Thus, it balances the merging and dividing steps efficiently compared to simpler methods. By maintaining a logarithmic growth with each split and merge step, merge sort tends to outperform algorithms like selection sort in scenarios involving large numbers of elements. This makes it a favorite for handling more significant sorting tasks in computer science.
Algorithm Efficiency
Algorithm efficiency is crucial when selecting which sorting technique to implement, particularly in the context of resource limitations and performance requirements. Efficiency is generally measured by the time and space complexity of the algorithm:
  • Time complexity refers to the rate at which the computational steps increase as the input size grows. For instance, an algorithm with \( O(n^2) \) time complexity will experience a quadratic increase in steps as more data is added.
  • Space complexity considers how much memory the algorithm requires relative to the input size.
In practical terms, more efficient algorithms (like merge sort with \( O(n \log n) \)) will handle larger datasets more effectively and quickly than less efficient ones (such as selection sort). Understanding these efficiencies helps developers make informed decisions about which algorithms will best suit their needs, balancing factors like the size of the input data, the processing power available, and the required speed of the operation.

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

(Bucket Sort) A bucket sort begins with a one-dimensional vector of positive integers to be sorted and a two-dimensional vector of integers with rows indexed from 0 to 9 and columns indexed from 0 to \(n 1\), where \(n\) is the number of values to be sorted. Each row of the two-dimensional vector is referred to as a bucket. Write a class named Bucketsort containing a function called sort that operates as follows: a. Place each value of the one-dimensional vector into a row of the bucket vector, based on the value's "ones" (rightmost) digit. For example, 97 is placed in row 7,3 is placed in row 3 and 100 is placed in row 0. This procedure is called a distribution pass. b. Loop through the bucket vector row by row, and copy the values back to the original vector. This procedure is called a gathering pass. The new order of the preceding values in the one-dimensional vector is 100,3 and 97 c. Repeat this process for each subsequent digit position (tens, hundreds, thousands, etc.). On the second (tens digit) pass, 100 is placed in row 0,3 is placed in row 0 (because 3 has no tens digit) and 97 is placed in row 9. After the gathering pass, the order of the values in the one-dimensional vector is 100,3 and \(97 .\) On the third (hundreds digit) pass, 100 is placed in row 1,3 is placed in row 0 and 97 is placed in row 0 (after the 3 ). After this last gathering pass, the original vector is in sorted order. Note that the two-dimensional vector of buckets is 10 times the length of the integer vector being sorted. This sorting technique provides better performance than a bubble sort, but requires much more memorythe bubble sort requires space for only one additional element of data. This comparison is an example of the spacetime trade-off: The bucket sort uses more memory than the bubble sort, but performs better. This version of the bucket sort requires copying all the data back to the original vector on each pass. Another possibility is to create a second two-dimensional bucket vector and repeatedly swap the data between the two bucket vectors.

(Bubble Sort) Implement bubble sortanother simple yet inefficient sorting technique. It is called bubble sort or sinking sort because smaller values gradually "bubble" their way to the top of the vector (i.e., toward the first element) like air bubbles rising in water, while the larger values sink to the bottom (end) of the vector. The technique uses nested loops to make several passes through the vector. Each pass compares successive pairs of elements. If a pair is in increasing order (or the values are equal), the bubble sort leaves the values as they are. If a pair is in decreasing order, the bubble sort swaps their values in the vector. The first pass compares the first two elements of the vector and swaps them if necessary. It then compares the second and third elements in the vector. The end of this pass compares the last two elements in the vector and swaps them if necessary. After one pass, the largest element will be in the last index. After two passes, the largest two elements will be in the last two indices. Explain why bubble sort is an \(O\left(n^{2}\right)\) algorithm.

In the text, we say that after the merge sort splits the vector into two subvectors, it then sorts these two subvectors and merges them. Why might someone be puzzled by our statement that "it then sorts these two subvectors"?

What key aspect of both the binary search and the merge sort accounts for the logarithmic portion of their respective Big Os?

(Quicksort) The recursive sorting technique called quicksort uses the following basic algorithm for a one-dimensional vector of values: a. Partitioning Step : Take the first element of the unsorted vector and determine its final location in the sorted vector (i.e., all values to the left of the element in the vector are less than the element, and all values to the right of the element in the vector are greater than the elementwe show how to do this below). We now have one element in its proper location and two unsorted subvectors. b. Recursion Step : Perform Step 1 on each unsorted subvector. Each time Step 1 is performed on a subvector, another element is placed in its final location of the sorted vector, and two unsorted subvectors are created. When a subvector consists of one element, that element is in its final location (because a one-element vector is already sorted). The basic algorithm seems simple enough, but how do we determine the final position of the first element of each subvector? As an example, consider the following set of values (the element in bold is the partitioning elementit will be placed in its final location in the sorted vector): $$\begin{array}{llllllllll} 37 & 2 & 6 & 4 & 89 & 8 & 10 & 12 & 68 & 45 \end{array}$$

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.

Sign-up for free