Question

(In Java) Do Shellsort twice, first using the suboptimal sequence for h which starts at half the length of the array and is halved for each pass. Then use Knuth’s interval sequence 1, 4, 13, 40 … for values of h.

For each sort, print the results in a table for:

a. How much time it took.

b. How many comparisons it made between two values.

Answer 1

am providing only sorting of elements code

public class ShellSort
{
private long[] item;

private int length;

public ShellSort(int max)
{
item = new long[max];
length = 0;
}

public void insert(long value)
{ item[length] = value;
length++;
}

public void display() {
System.out.print("Item:");
for (int j = 0; j < length; j++)
System.out.print(item[j] + " ");
System.out.println("");
}

public void shellSort() {
int inner, outer;
long temp;

int h = 1;
while (h <= length / 3)
h = h * 3 + 1;

while (h > 0)
{
  
for (outer = h; outer < length; outer++) {
temp = data[outer];
inner = outer;
  
while (inner > h - 1 && item[inner - h] >= temp) {
item[inner] = item[inner - h];
inner -= h;
}
item[inner] = temp;
}
h = (h - 1) / 3;
}
}

public static void main(String[] args)
{
int maxSize = 10;
ShellSort arr = new ShellSort(maxSize);

for (int j = 0; j < maxSize; j++) {
long n = (int) (java.lang.Math.random() * 99);
arr.insert(n);
}
arr.display();
arr.shellSort();
arr.display();
}
}