Question

Consider the following list: \(\begin{array}{lllllllllll}2 & 10 & 17 & 45 & 49 & 55 & 68 & 85 & 92 & 98 & 110\end{array}\) Using the binary search, how many comparisons are required to determine whether the following items are in the list or not? Show the values of first, last, and middle and the number of comparisons after each iteration of the loop. a. 15 b. 49 c. 98 d. 99

Short answer

15 (4 comparisons, not found); 49 (4 comparisons, found); 98 (3 comparisons, found); 99 (4 comparisons, not found).

Step-by-step solution

Understanding Binary Search

Binary search is a technique used to find an element in a sorted array. It works by repeatedly dividing the search interval in half. If the value of the search key is less than the item in the middle of the interval, we narrow the interval to the lower half. Otherwise, we narrow it to the upper half. We continue this process until the desired element is found or the interval is empty.

Initial Setup for Binary Search

For each element search, we must set: - `first` to 0 (the first index of the array), - `last` to 10 (the last index of the array), - `middle` is calculated as the average of `first` and `last`, using integer division.

Search for Element 15

Initial values: `first = 0`, `last = 10` 1. Compute `middle = (0 + 10) // 2 = 5`. Compare list[5] = 55 with 15. Since 15 < 55, set `last = 4`. 2. Compute `middle = (0 + 4) // 2 = 2`. Compare list[2] = 17 with 15. Since 15 < 17, set `last = 1`. 3. Compute `middle = (0 + 1) // 2 = 0`. Compare list[0] = 2 with 15. Since 15 > 2, set `first = 1`. 4. Compute `middle = (1 + 1) // 2 = 1`. Compare list[1] = 10 with 15. Since 15 > 10, set `first = 2`. Since `first > last`, the element 15 is not in the list. Total comparisons: 4.

Search for Element 49

Initial values: `first = 0`, `last = 10` 1. Compute `middle = (0 + 10) // 2 = 5`. Compare list[5] = 55 with 49. Since 49 < 55, set `last = 4`. 2. Compute `middle = (0 + 4) // 2 = 2`. Compare list[2] = 17 with 49. Since 49 > 17, set `first = 3`. 3. Compute `middle = (3 + 4) // 2 = 3`. Compare list[3] = 45 with 49. Since 49 > 45, set `first = 4`. 4. Compute `middle = (4 + 4) // 2 = 4`. Compare list[4] = 49 with 49. They are equal. The element 49 is found. Total comparisons: 4.

Search for Element 98

Initial values: `first = 0`, `last = 10` 1. Compute `middle = (0 + 10) // 2 = 5`. Compare list[5] = 55 with 98. Since 98 > 55, set `first = 6`. 2. Compute `middle = (6 + 10) // 2 = 8`. Compare list[8] = 92 with 98. Since 98 > 92, set `first = 9`. 3. Compute `middle = (9 + 10) // 2 = 9`. Compare list[9] = 98 with 98. They are equal. The element 98 is found. Total comparisons: 3.

Search for Element 99

Initial values: `first = 0`, `last = 10` 1. Compute `middle = (0 + 10) // 2 = 5`. Compare list[5] = 55 with 99. Since 99 > 55, set `first = 6`. 2. Compute `middle = (6 + 10) // 2 = 8`. Compare list[8] = 92 with 99. Since 99 > 92, set `first = 9`. 3. Compute `middle = (9 + 10) // 2 = 9`. Compare list[9] = 98 with 99. Since 99 > 98, set `first = 10`. 4. Compute `middle = (10 + 10) // 2 = 10`. Compare list[10] = 110 with 99. Since 99 < 110, set `last = 9`. Since `first > last`, the element 99 is not in the list. Total comparisons: 4.

Key concepts

Algorithm Analysis

Binary search is a foundational algorithm in computer science, used to efficiently find an item in a sorted array. The core idea is to repeatedly divide the search space in half. Initially, you determine the size of the array and set two pointers, `first` and `last`, to the beginning and end of the array, respectively. During each iteration:

• Calculate the `middle` index from `first` and `last`.
• Compare the middle element to the target.
• If the middle element is the target, the search ends successfully.
• If the target is smaller, adjust `last` to just before `middle`.
• If the target is larger, move `first` to just past the `middle`.

This analysis helps in understanding the efficiency of the algorithm, particularly how it optimizes searching by utilizing the characteristics of sorted data. Understanding each adjustment of `first`, `last`, and `middle` for every comparison is crucial to mastering binary search.

Search Optimization

Search optimization refers to how efficiently an algorithm finds a target within a dataset. Binary search is an excellent example of an optimized searching method, especially suited for sorted data. Its efficiency comes from the fact that, rather than scanning each element individually, it quickly narrows down the potential location of the target by focusing on only half of the current section every time.
For example, when searching for the number 49 in the given array, binary search efficiently zeroes in on the correct part of the data. After comparing to four different elements, it successfully finds the number. This is drastically faster than a linear search, which might need to check every single element sequentially. Thus, understanding binary search highlights the power of search optimization in improving computational tasks.

Computational Complexity

The concept of computational complexity is critical in evaluating how an algorithm's resource requirements grow as the input size increases. Binary search operates with a complexity of \( O(\log n) \) in terms of time. This is because each comparison cuts the search space in half. For an array of size \( n \), you will at most perform \( \log_2(n) \) comparisons.
In practical terms, for the array provided, with a size of 11, this means the worst-case scenario could involve around 4 comparisons. This efficiency is what makes binary search a preferable choice for large datasets, compared to linear search, which has a complexity of \( O(n) \). Understanding these complexities allows one to choose the correct algorithm for a task depending on the size and nature of the data involved.

Sorted Arrays

Sorted arrays are sequences of data organized in a specific order, such as ascending or descending. Binary search requires that the data it operates on be sorted. This is because the assumption that allows the search space to be halved repeatedly would not hold if the elements were unordered.
In our exercise, the array is already sorted in ascending order, which facilitates the binary search process. Without this sorting, there would be no systematic way to eliminate half of the elements from consideration during each comparison. Therefore, whenever considering binary search, or similar algorithms, ensuring your data is sorted is a prerequisite to achieving the optimal efficiency binary search is known for.