选择排序和冒泡排序之间的比较数量保持不变

Question

I have written this program to compare the number of operations needed to sort a random numbers using both selection and bubble sort. However, these numbers keep coming up the same and I can't figure out where my code went wrong.

static int num_comps;   

public static void main(String[] args) 
{
    Random rnd = new Random();

    // max size of array
    // number of N inputs
    int array_size = 32768;
    int num_datasets = 12;

    // allocate array once to the max size
    int[] vals = new int[array_size];

    // temp array with allocated array to max size
    int[] tvals = new int[array_size];

    // array to hold operation counts
    int[] op_counts = new int[num_datasets];
    int[] op_counts2 = new int[num_datasets];

    // array to hold the size of each array
    //
    int[] arraySizes = new int[num_datasets];

    int i;
    int j;
    int sz;

    for (i = 0, sz = 16; i < num_datasets; i++, sz *= 2)
        arraySizes[i] = sz;

    for (int iter = 0; iter < num_datasets; iter++)
    {
        int curr_size = arraySizes[iter];

        // load array with random values
        //
        for (i = 0; i < curr_size; i++)
            vals[i] = rnd.nextInt(4999);

        for (i = 0; i < curr_size; i++)
            tvals[i] = vals[i];

        // run the bubble sort algorithm
        //
        num_comps = 0;
        bubbleSort(tvals, curr_size);
        op_counts[iter] = num_comps;
        //System.out.println("Num comps at " + iter + " is " + num_comps);

        // run the selection-sort algorithm
        num_comps = 0;
        selectionSort(tvals, curr_size);
        op_counts2[iter] = num_comps;
        //System.out.println("Num comps at " + iter + " is " + num_comps);
    }

    System.out.println("Operation Counts (N vs. op Count): ");
    for (int k = 0; k < num_datasets; k++)
        System.out.println(arraySizes[k] + "\t\t" + op_counts[k] + "\t\t" + op_counts2[k]);
}

static void bubbleSort(int vals[], int curr_size)
{
    int temp;
    for (int i = 0; i < curr_size - 1; i++)
    {
        for (int j = 0; j < curr_size - i - 1; j++)
        {
            // swap
            num_comps = num_comps + 1;
            if (vals[j+1] < vals[j])
            {
                temp = vals[j];
                vals[j] = vals[j+1];
                vals[j+1] = temp;
            }
        }
    }
}

static void selectionSort(int vals[], int curr_size)
{
    int temp;
    for(int i=0; i<curr_size - 1; i++)
     {
        for(int j=i+1; j<curr_size; j++)
        {
            num_comps = num_comps + 1;
            if(vals[i] > vals[j] )
            {
                temp = vals[j];
                vals[j] = vals[i];
                vals[i] = temp;
            }
        }
     }
}

Answer 1

Your selection sort algorithm does not search for the lowest value in the list. And swapping it afterwards with the index of the outer loop.

You should do something like this:

static void selectionSort(int vals[], int curr_size)
{
    int temp;
    for(int i=0; i<curr_size - 1; i++)
    {
        int lowest = i;
        for(int j=i+1; j<curr_size; j++)
        {
            num_comps = num_comps + 1;
            if(vals[lowest] > vals[j] )
            {
                lowest = j;
            }
        }

        // swap lowest with current index
        temp = vals[lowest];
        vals[lowest] = vals[i];
        vals[i] = temp;
     }
}

(of course this can be optimized further) The strength of this algorithm is not the amount of of comparisons, but this amount of swaps (which is at a minimum, I suppose).

Your bubble sort algorithm seems ok to me.

Both have the same two loops, so comparing the counts of the current implementations indeed result in the same values. But, I think you can optimize the bubble sort, to stop earlier (when no swaps were found). Again, the strength of sorting algorithms depends on the used ones, and are not necessarily the least amount of comparisons. So Using the correct algorithm for your specific task, and thereby circumventing the task-specific high cost operations, is important!