Posted: August 24th, 2022

Programming Assignment

( Insertion Sort )

This is a simple programming assignment to implement insertion sort algorithm and to observe its worst- case, best-case, and average-case performance. The performance measurement is in terms of the number of key-comparisons, rather than the actual running time.

Implement insertion-sort algorithm without use of recursion. (A recursive implementation of insertion sort for large size n may cause run-time stack overflow.) To keep track of the number of key-comparisons, it is recommended that the sorting algorithm makes use of a Boolean function SMALLER(A, i, j) to do the following:

· Increment a global counter, COMPCOUNT, to keep track of the number of key-comparisons per- formed by the algorithm. (This count is initialized to 0 at the beginning of the algorithm.)

· Perform a comparison between A[i] and A[j]. Return TRUE if A[i] < A[j]. Otherwise, return FALSE. Carry out the following experiments.

1 Small-Size Array, n = 32.

Run the algorithm for n = 32 and for each of the following cases:

(1) Worst-case data input; (2) Best-case data input; (3) Random data input. (Performance on random data represents average-case.)

For each case, print n, input array, output array (sorted data), and the number of key-comparisons. Does the number of key-comparisons agree with the theoretical values? Theoretically, the worst-cse number of key comparisons is (n2 − n)/2, and the average number is (n2 − n)/4, which is half of the worst-case.

2 Increasing Array Sizes, n = 100, n = 1000, n = 10000.

Run the algorithm for each of these increasing array sizes and for random data input. For each case, print n and the resulting number of key-comparisons. (Note that for large n, it is not practical to print the actual input/output arrays! Also, since the algorithm has O(n2) time complexity, an array size larger than 10000 may not be practical.)

( − )Does the number of key-comparisons show O(n2) performance? That is, when the array size is increased by a factor of 10, does the number of operations (comparisons) increase by approximately a factor of 100? What is the constant factor for the O(n2) performance? Note: Theoretically, the average number of key-comparisons for insertion sort is (n2 n)/4. Therefore, for large n, the number of comparisons should be approximately n2/4.

Your program must be in C, C++, or JAVA. Submit your program on Canvas as followd:

1. The source code of your program. (The TA needs to visually read your program to evaluate it, and also run the program to verify that it works.)

( 1 )

2. The output as produced by your program.

3. A short discussion of the results, tabulating the results, and comparing them with the theoretical values.

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(Sort by Insertion)

This is a short programming assignment in which you will develop the insertion sort algorithm and evaluate its worst-case, best-case, and average-case performance. The number of key-comparisons is used to measure performance rather than the actual running time.

Implement an insertion-sort method that does not rely on recursion. (For high n, a recursive implementation of insertion sort may result in run-time stack overflow.) It is advised that the sorting algorithm employ the Boolean function SMALLER(A, I j) to do the following to keep track of the number of key-comparisons:

Increase the value of a global counter, COMPCOUNT, to keep track of the number of key-comparisons performed by the algorithm. (This count is initialized to 0 at the beginning of the algorithm.)

· Perform a comparison between A[i

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