پروژه برنامه نویسی و طراحی وب سایت

First Pass:  d = (6 + 1) / 2 = 3.  Compare 1st and 4th , 2nd and 5th, and 3rd and 6th items since they are  3 positions away from each other))
Second Pass:  value for d is halved  d = (3 + 1) / 2 = 2.  Compare items two places away such as 1st and 3rd
Third Pass:  value for d is halved  d = (2 + 1) / 2 = 1.  Compare items one place away such as 1st and 2nd ..
Last Pass:  sort continues until d = 1 and the pass occurs without any swaps.

This sorting process, with its comparison model, is an efficient sorting algorithm.

//Shell Sort Function for Descending Order
void ShellSort( apvector &num)
{
     int i, temp, flag = 1, numLength = num.length( );
     int d = numLength;
     while( flag || (d > 1))      // boolean flag (true when not equal to 0)
     {
          flag = 0;           // reset flag to 0 to check for future swaps
          d = (d+1) / 2;
          for (i = 0; i < (numLength - d); i++)
        {
               if (num[i + d] > num[i])
              {
                      temp = num[i + d];      // swap positions i+d and i
                      num[i + d] = num[i];
                      num[i] = temp;
                      flag = 1;                  // tells swap has occurred
              }
         }
     }
     return;