Why do we make n-1 iterations in bubble sort algorithm

Question

Most common way of bubble sort algorithm is to have two for loops. Inner one being done from j=0 until j n-i-1. I assume we substract minus i, because when we reach last element we don't compare it because we don't have an element after him. But why do we use n-1. Why we don't run outer loop from i=0 until i < n and inner from j=0 until n-i? Could someone explain it to me, tutorials on internet does not emphasize this.

for (int i = 0; i < n - 1; i++) // Why do we have n-1 here?
    {
        swapped = false;
        for (int j = 0; j < n - i - 1; j++)
        {
            countComparisons++;
            if (arr[j] > arr[j + 1])
            {
                countSwaps++;
                swap(&arr[j], &arr[j + 1]);
                swapped = true;
            }

        }
     }

For example, if I have an array with 6 elements, why do I only need to make 5 iterations?

Answer 1

Because a swap requires at least two elements.

So if you have 6 elements, you only need to consider 5 consecutive pairs.

Answer 2

For comparison purposes in an array, two adjacent cells are needed; in an array of 6 elements, you do 5 comparisons only; in an array of 10 elements, 9 comparisons, and so on:

array and comparisons between adjacent cells

So for 7 elements, just 6 comparisons are done, hence the general rule of n-1 in the outer for loop

About the n-1-i expression, remember that the highest (or lowest, depending on the ordering criterion) value in the bubble sort goes to the last position in the array after the first cycle, so there is no need to compare that value with anything else, therefore the array has to be "shortened" 1 cell at a time, and the value of i in the outer loop is the counter responsible for that in the inner loop:

5 | 3 | 9 | 20 | elements (n) = 4

after first cycle (i = 0), 20 has reached its correct position within the array (using an ascending order), leaving us with an array of 3 elements to do comparisons to; in next cycle, i will be equal to 1, and as n-1 remains the same, we need to substract 1 in that expression to "shorten" the array: n-1-i = 4-1-1 = 2, which is the index of the last element in that new array as well as the quantity of comparisons needed.

Hope it helps!