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

Short Answer

Expert verified
The logarithmic part comes from repeatedly halving the dataset in both algorithms.

Step by step solution

01

Understand Binary Search

Binary search is a search algorithm that efficiently finds the position of a target value within a sorted array. It works by dividing the dataset in half with each step, thereby reducing the problem size logarithmically. The key to its logarithmic time complexity, O(log n), lies in this consistent halving of the dataset until the target value is found or the subset is reduced to zero.
02

Understand Merge Sort

Merge sort is a divide-and-conquer algorithm that sorts an array by recursively dividing it into two halves, sorting each half, and then merging the sorted halves back together. The key aspect is the recursive division of the array, leading to a depth of recursion that is logarithmic in relation to the size of the array. Thus, the logarithmic component of its time complexity, O(n log n), arises from this division.
03

Compare and Generalize

Both algorithms involve the consistent division of datasets into halves, which represents logarithmic behavior. For binary search, this division step is the primary recursive call. For merge sort, it's the recursive splitting and merging process. In both, the logarithmic aspect (log base 2, specifically) measures how many times a dataset can be divided in half until it is broken down completely.
04

Conclusion

The logarithmic portion in the Big O notation of both algorithms (O(log n) for binary search and O(n log n) for merge sort) arises from the repeated halving of the data structure size. These divisions inherently follow a logarithmic scale because the extent of division steps required corresponds to the power of 2 that approximately equals the size of the dataset.

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Key Concepts

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

Binary Search
Binary search is an efficient algorithm used to find the position of a target value within a sorted array. Imagine you are tasked with searching for a name in a sorted phonebook. Instead of scanning page by page, you open the book to the halfway point, determine whether the name, if present, would be on the left or right half, then discard the irrelevant half.

By reducing the search space by half at each step, binary search exemplifies a powerful concept in algorithm design. This reduction allows the search algorithm to quickly narrow down the potential locations of the target.

For example, if you start with 32 elements, one search operation reduces the problem space to 16, then to 8, 4, 2, and ultimately just 1 element.

Hence, each check pushes the process forward in a logically calculated way, following a pattern that mirrors the properties of a logarithm, where the time complexity becomes \( \mathcal{O}(\log n) \). It's essential to note that logarithm, in this case, generally uses base 2, denoting how many times the dataset is divided.
Merge Sort
Merge sort is a classic example of a divide-and-conquer algorithm. Think of sorting a pile of shuffled papers. Instead of sorting them all at once, you divide the pile into smaller sections, sort each section independently, and then merge the sections back together.

This algorithm effectively breaks down the task of sorting a large array into smaller and more manageable tasks. Specifically, it divides the array into two halves, sorts those recursively, and then carefully merges them, preserving the order.

The key to understanding merge sort's efficiency lies in its recursive nature. The depth of this recursion tree is logarithmic to the size of the array, which implies that it divides the problem swiftly.

Thus, the complexity includes the linear time needed to merge the sorted arrays at each level of recursion, ultimately resulting in a time complexity of \( \mathcal{O}(n \log n) \). This merging process effectively captures the logarithmic aspect, stemming from repeated halving at each layer of recursion.
Logarithmic Complexity
When we discuss logarithmic complexity, we talk about operations or processes that become significantly faster with each step, simply because they decrease the size of the problem by a factor proportional to logarithms. This concept is crucial in computer science, particularly in efficient algorithm design.

A linear search through a list of items might analyze each element one after the other, resulting in an \( \mathcal{O}(n) \) time complexity. Contrastingly, logarithmic algorithms like binary search or log components in merge sort speed up the process by cutting down the problem size decisively with each operation.

Consider a dataset of size \( n \). The number of times this dataset can be halved until reaching a single element is exactly the logarithm to base 2 of \( n \). Hence, with each division, the task becomes smaller and easier to manage.

This logarithmic behavior is not just limited to search or sorting algorithms. Many structures and paradigms such as balanced trees and certain dynamic programming approaches also embody this. The utility of logarithmic complexity, in essence, lies in how it transforms seemingly large problems into digestible and palatable pieces for computation, highlighting efficiency and scalability.

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Most popular questions from this chapter

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 ________.

(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.

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"?

(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.

(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): $$\begin{array}{llllllllll} 37 & 2 & 6 & 4 & 89 & 8 & 10 & 12 & 68 & 45 \end{array}$$

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