Question

1. Implement the merge-sort algorithm

a. Determine the correct basic operation

2. Test your algorithm on inputs of size 2^10, 2^11 ... 2^20.

If this is too slow then just run it on any set of powers of two.

3. Show the running time of each of these inputs

Choose a basic operation

Count the total number times that this basic operation is done for each

power of 2.

Display this count in a table

Answer 1

Since you have not mentioned the language of your preference, I am providing the code in JAVA,

CODE

import java.util.Random;

import java.util.Scanner;

public class Main

{

public static void merge(int arr[], int l, int m, int r)

{

int n1 = m - l + 1;

int n2 = r - m;

int L[] = new int [n1];

int R[] = new int [n2];

for (int i=0; i<n1; ++i)

L[i] = arr[l + i];

for (int j=0; j<n2; ++j)

R[j] = arr[m + 1+ j];

int i = 0, j = 0;

int k = l;

while (i < n1 && j < n2)

{

if (L[i] <= R[j])

{

arr[k] = L[i];

i++;

}

else

{

arr[k] = R[j];

j++;

}

k++;

}

while (i < n1)

{

arr[k] = L[i];

i++;

k++;

}

while (j < n2)

{

arr[k] = R[j];

j++;

k++;

}

}

public static void mergeSort(int arr[], int l, int r)

{

if (l < r)

{

int m = (l+r)/2;

mergeSort(arr, l, m);

mergeSort(arr , m+1, r);

merge(arr, l, m, r);

}

}

public static void printArray(int arr[]) {

int n = arr.length;

for (int i=0; i<n; ++i)

System.out.print(arr[i] + " ");

System.out.println();

}

public static void main(String args[]) {

Scanner sc = new Scanner(System.in);

for (int k=10; k<=20; k++) {

int n = (int) Math.pow(2, k);

int a[] = new int[n];

int temp[] = new int[n];

Random rand = new Random();

for(int i=0; i<n; i++) {

a[i] = rand.nextInt(100);

}

for(int i=0; i<n; i++) {

temp[i] = a[i];

}

long startTime, endTime, duration;

a = temp;

startTime = System.nanoTime();

mergeSort(a, 0, a.length-1);

endTime = System.nanoTime();

duration = (endTime - startTime) / 1000000;

System.out.println("For n = " + n + ", Merge Sort took " + duration + " milliseconds\n");

}

sc.close();

}

}