Part Algorithm Analysis And Searching Algorithms

Part Valgorithm Analysisconsider Searching Algorithms On The Following

Part Valgorithm Analysisconsider Searching Algorithms On The Following

Part V Algorithm Analysis Consider searching algorithms on the following array of data: [] Suppose you want to implement a searching algorithm to see if the data set contains the number 19. Demonstrate how the search would go if you used: A sequential search A binary search State the runtime for each of the searches, in this example, and for general data sets of size n . Address the issue of the order of the data in binary searching. Suppose an algorithm that processes a data set of size 8 has a runtime of 72. The same algorithm has a runtime of 110 when applied to a data set of size 10; and when applied to a data set of size 20, it has a runtime of 420. Using big-O notation, state the runtime for this algorithm for the general case of a data set of size n . Suppose you develop an algorithm that processes the first element of an array (length of n ), then processes the first 2 elements, then the first 3 elements, and so on, until the last iteration of a loop, when it processes all elements. Thus, if n = 4, the runtime would be 1 + 2 + 3 + 4 = 10. Create a table that depicts the runtime for arrays of length 1 to 10. Would you expect the general runtime to be O( n ), O( n 2 ), O( n 3 ), or some other function of n ? Explain

Paper For Above instruction

Introduction

Searching algorithms are fundamental tools in computer science, enabling efficient data retrieval from large datasets. The effectiveness of a search depends on the properties of the algorithm, such as whether the data is ordered, and the size of the dataset. This paper analyzes different searching algorithms, examines their runtimes under specific conditions, and explores the complexity of iterative summation runtimes in relation to data size.

Sequential Search and Binary Search

Sequential search, also known as linear search, involves checking each element of a dataset sequentially until the target value is found or the list ends. Suppose the dataset is an unsorted array and we want to find whether the number 19 exists within it. If the data set is [3, 7, 19, 25, 30], the sequential search compares '19' with each element in order: 3, 7, and then 19. It would require three comparisons, and since the worst-case scenario involves traversing the entire array (if 19 is not present or at the end), the time complexity is O(n), where n is the size of the dataset. In general, for a dataset of size n, the sequential search runs in linear time, O(n).

In contrast, binary search is efficient but requires that the data is sorted prior to the search. For example, in an ordered array [3, 7, 19, 25, 30], binary search repeatedly divides the search interval in half, comparing the target to the middle element. To find 19, the algorithm compares against the middle element—say, 19 at index 2 (zero-based indexing)—and finds it immediately if the dataset is sorted accordingly. The process involves log2(n) comparisons, so the runtime is O(log n). The key condition for binary search is that the data must be sorted; otherwise, binary search cannot reliably locate elements.

Analysis of Given Runtime Data

The provided runtime data for the algorithm applied to various dataset sizes presents a pattern:

- Size 8: runtime 72

- Size 10: runtime 110

- Size 20: runtime 420

To analyze this, we consider that the runtime appears to grow faster than linear but less than quadratic. Fitting these data points to a polynomial model suggests the runtime could be proportional to n^b, where b is between 1 and 2.

Using approximation:

- For n=8, runtime=72

- For n=10, runtime=110

- For n=20, runtime=420

Calculating ratios:

- 110/72 ≈ 1.527

- 420/110 ≈ 3.818

Similarly, the ratio of sizes:

- 10/8=1.25

- 20/10=2

If runtime ~ kn^b, then:

- 72 ≈ k*8^b

- 110 ≈ k*10^b

- 420 ≈ k*20^b

Dividing the equations:

- 110/72 ≈ (10/8)^b → 1.527 ≈ (1.25)^b → b ≈ log(1.527)/log(1.25) ≈ 0.184/0.097 ≈ 1.89

- 420/110 ≈ (20/10)^b → 3.818 ≈ 2^b → b ≈ log(3.818)/log(2) ≈ 0.582/0.301 ≈ 1.93

Since both estimates hover near 2, the runtime appears to be proportional to n^2, or O(n^2).

Runtime Complexity in Big-O Notation

Considering the fitted data and the growth rate, the algorithm's runtime can be expressed as O(n^2) in the general case. This indicates that as the dataset size doubles, the runtime increases approximately four times, consistent with quadratic complexity. Such growth suggests the algorithm might involve nested iterations or other quadratic operations, which are characteristic of quadratic time complexity.

Iterative Summation Runtime Analysis

The described algorithm computes the sum 1 + 2 + 3 + ... + n, which results in a sum that is quadratic in nature: n(n+1)/2. When tabulated for n from 1 to 10, the total runtime is as follows:

This sequence aligns with the formula n(n+1)/2, which indicates the total runtime is proportional to n^2. Therefore, the overall complexity of this sum operation is O(n^2). As n increases, the runtime grows quadratically, which is consistent with the mathematical sum of an arithmetic series.

Conclusion

In conclusion, searching algorithms vary significantly in their efficiency based on data organization. Sequential search operates in linear time, while binary search benefits from logarithmic time complexity, provided the data is sorted. The analysis of specific runtime data indicates an O(n^2) complexity for the algorithm in question, likely involving nested iterations. The summation approach demonstrates quadratic growth, reinforcing the importance of algorithmic efficiency considerations for large datasets. Selecting the appropriate algorithm depends on data characteristics and the required performance constraints in practical applications.

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