Optimal bubble sorting algorithm for an array of arrays of numbers

Question

Fix positive integers n and k.

Let A be an array of length n with A[i] an array of length k

Answer 1

This is not an optimal answer, but i would like to share my attempt as someone may improve it. I did not thought about finding a formula to calculate the minimum number of swaps but rather on the optimal algorithm. The algorithm is based on k = 2.

The basic idea is based on information gain. Let us assume that A = {[i,j] : 1<=i<=n, 1<=j<=n} represents a configuration. In each step, we have 4 * (n-1) possible swapping to move from one configuration to another configuration. For example if n = 2 (i.e. A = [ {2,2}, {1,1} ] ), then we have 4 possible swapping A[0][0] <-> A[1][0], A[0][0] <-> A[1][1], A[0][1] <-> A[1][0], and A[0][1] <-> A[1][1]. Thus, our objective is to select the swap that has high information gain when we need to move from one configuration to another configuration.

The tricky part will be "how to calculate the information gain". In my solution (below), the information gain is based on the distance of a value from its correct position. Let me show you my code (written in C++) to understand what i am trying to say:

const int n = 5;
const int k = 2;

int gain(int item, int from, int to)
{
    if (to > from)
        return item - to;
    else
        return to - item ;
}

void swap(int &x, int &y)
{
    int temp = x;
    x = y;
    y = temp;
}

void print_config (int A[][k])
{
    cout << "[";
    for (int i=0; i max_gain) {
                   r = i;
                   c = j;
                   max_gain = G[i][j];
                }

        // Did we gain more information. If not terminate
        if (max_gain < 0) break;

        switch (c)
        {
            case 0: swap(A[r][0], A[r+1][0]); break;
            case 1: swap(A[r][0], A[r+1][1]); break;
            case 2: swap(A[r][1], A[r+1][0]); break;
            case 3: swap(A[r][1], A[r+1][1]); break;
        }

        print_config(A);
        num_swaps++;

    } while (1);
    cout << "Number of swaps is " << num_swaps << endl;
}

I ran the above code for cases n=1,2,... and 7. Here are the answers (number of swaps) respectively: 0, 2, 5, 10, 15, 23 (very close), and 31. I think that the function gain() does not work well when n is even. Can you confirm that by validating the number of swaps when n = 7. The lower bound of your equation is 31 so this is the optimal number of swaps when n = 7.

I am printing here the output when n = 5 (since you are looking for a pattern):

[ [5, 5],  [4, 4],  [3, 3],  [2, 2],  [1, 1] ]
[ [4, 5],  [5, 4],  [3, 3],  [2, 2],  [1, 1] ]
[ [4, 5],  [3, 4],  [5, 3],  [2, 2],  [1, 1] ]
[ [4, 5],  [3, 4],  [2, 3],  [5, 2],  [1, 1] ]
[ [4, 5],  [3, 4],  [2, 3],  [1, 2],  [5, 1] ]
[ [4, 3],  [5, 4],  [2, 3],  [1, 2],  [5, 1] ]
[ [4, 3],  [2, 4],  [5, 3],  [1, 2],  [5, 1] ]
[ [4, 3],  [2, 4],  [1, 3],  [5, 2],  [5, 1] ]
[ [4, 3],  [2, 4],  [1, 3],  [1, 2],  [5, 5] ]
[ [4, 3],  [2, 1],  [4, 3],  [1, 2],  [5, 5] ]
[ [1, 3],  [2, 4],  [4, 3],  [1, 2],  [5, 5] ]
[ [1, 3],  [2, 4],  [1, 3],  [4, 2],  [5, 5] ]
[ [1, 3],  [2, 1],  [4, 3],  [4, 2],  [5, 5] ]
[ [1, 1],  [2, 3],  [4, 3],  [4, 2],  [5, 5] ]
[ [1, 1],  [2, 3],  [2, 3],  [4, 4],  [5, 5] ]
[ [1, 1],  [2, 2],  [3, 3],  [4, 4],  [5, 5] ]