Question

suppose I have an array of 10 ints, and I’m using binary search to find a number, let’s take for example numbers

1 2 3 4 5 6 7 8 9 10

and I’m using this method

static void binarySearch(int n, int[] a, int low, int high)
    {
        int mid = (high + low) / 2;
        if(low > high)
            System.out.println(n+" was not found after "+counter+" comparisons");
        else if(a[mid] == n)
        {
            counter++;
            System.out.println(n+" was found at position "+mid+" after "+counter+" comparisons");
        }            
        else if(a[mid] < n)
        {
            counter++;
            binarySearch(n, a, mid+1, high);
        }            
        else
        {
            counter++;
            binarySearch(n, a, low, mid-1);
        }            
    }

what is the proper way of calling the method binarySearch(5, a, 0, a.lenght)
or
binarySearch(5, a, 0, a.lenght-1)

I know they both will find the number, but they will find it at different index; thus taking more comparison

Answer 1

Well let’s do some tests shall we?

First, let’s search for each number in the array. We get:

binarySearch(i, array, 0, array.length);

1 was found at position 0 after 3 comparisons
2 was found at position 1 after 4 comparisons
3 was found at position 2 after 2 comparisons
4 was found at position 3 after 3 comparisons
5 was found at position 4 after 4 comparisons
6 was found at position 5 after 1 comparisons
7 was found at position 6 after 3 comparisons
8 was found at position 7 after 4 comparisons
9 was found at position 8 after 2 comparisons
10 was found at position 9 after 3 comparisons
Average: 2.9 comparisons

binarySearch(i, array, 0, array.length - 1);

1 was found at position 0 after 3 comparisons
2 was found at position 1 after 2 comparisons
3 was found at position 2 after 3 comparisons
4 was found at position 3 after 4 comparisons
5 was found at position 4 after 1 comparisons
6 was found at position 5 after 3 comparisons
7 was found at position 6 after 4 comparisons
8 was found at position 7 after 2 comparisons
9 was found at position 8 after 3 comparisons
10 was found at position 9 after 4 comparisons
Average: 2.9 comparisons

As you can see, variances do appear, but the average remains constant.
Now let’s test for bigger numbers:

100000 items
binarySearch(i, array, 0, array.length);
Average: 15.68946 comparisons
binarySearch(i, array, 0, array.length - 1);
Average: 15.68946 comparisons

200000 items
binarySearch(i, array, 0, array.length);
Average: 16.689375 comparisons
binarySearch(i, array, 0, array.length - 1);
Average: 16.689375 comparisons

500000 items
binarySearch(i, array, 0, array.length);
Average: 17.951464 comparisons
binarySearch(i, array, 0, array.length - 1);
Average: 17.951464 comparisons

Hence, on average it doesn’t go either way. For the sake of convention I would advise using the exclusive upper bound version: binarySearch(i, array, 0, array.length);