Write a computer program that prompts the user for one number, n for the number of items in the array to sort, and create and sort 1000 different arrays of this size timing the run to get an average time to sort an array of this size.

Module 7 showed that one way of comparing different algorithms for accomplishing the same task is complexity analysis. You will recall that in complexity analysis we express the time an algorithm takes to run as a function of the size of the input, and we used the big-Oh notation. For example, if an algorithm has a complexity of O(1), then it always runs in the same amount of time, no matter what the size of the input is; if it O(n), then the time it takes for the algorithm to run is proportional to the size of the input. However, complexity analysis has a number of limitations. For example, big-Oh analysis concerns the worst case scenario. For example, some sorting algorithms with a complexity of O(n^2) often run considerably faster if the list that it receives as input is (almost) sorted; other sorting algorithms with a complexity of O(n^2) always take the same amount of time, no matter what state the list is in. Also, in big-Oh, we look at the dominant term in our calculation of the complexity of the algorithm. Thus, when we analyze an algorithm and discover that it runs in 75,312 + n time units, we still say that it has a complexity of O(n). It is therefore deemed to be better than an algorithm that runs in .007 + n^2 time units, as this algorithm has a complexity of O(n^2).

We also saw the rationale behind this: If n becomes sufficiently large, the other factors become insignificant. Fortunately, there is another way to determine how long it takes for an algorithm to run, namely timing experiments. In a timing experiment, you actually implement the algorithm in a programming language, such as Java or C++, and simply measure how long it takes for the algorithm to run. In the term project for this course, you are going to conduct a timing experiment and compare the results with the results you would get from a complexity analysis. We will compare Bubble Sort with Selection Sort. In its least sophisticated form, bubble sort (http://en.wikipedia.org/wiki/Bubble_sort) works as follows: Assuming that the list contains n elements. Compare the first and the second element in the list, and swap them if the last element is smaller than the preceding one; otherwise, do nothing to this pair. Now, compare the second and third elements and swap them if the first of them is larger than the second; otherwise, do nothing to this pair. Move on the next pair and continue the process until you reach the end of the list.

A little reflection will show that at the end of this iteration, the last element in the list is now the largest element in the list. The last element has bubbled to the top. Now repeat the process but rather than going to the end of the list, stop when you reach n-1. Now repeat the process again, but rather than going to the end of the list, stop when you reach n-2. Keep repeating this until you reach 1. The Wikipedia entry has a little simulation that shows how bubble sort works. The code looks something like: bubbleSort(array A){ n = length(A); for(j = n; j > 0, j–) for(i = 1; i < j; i++) { if A[i-1] > A[i] swap(A,i-1, i); } } } swap obviously swaps the elements and can be defined as: swap(A, pos1, pos2) { temp = A[pos1]; A[pos1] = A[pos2]; A[pos2] = temp; } Another sort is selection sort (http://en.wikipedia.org/wiki/Selection_sort). We saw selection sort in the question in the sub-module on how to determine the complexity of an algorithm. Array A contains n elements, the elements to be sorted. The algorithm starts at the first position in the array and looks through the array for the smallest element. Once it reaches the end of the array, it puts that element in the first cell of the array. It then restarts the whole process from the second position in the array, and continues until the entire array has been sorted. selection_sort(array A) { int i,j int iMin; for(j = 0; j < n; j++){ iMin = j; for ( i = j+1; i < n; i++) { if (a[i] < a[iMin]) { iMin = i; } } if ( iMin != j ) { swap(a[j], a[iMin]); } } }

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