Question

I’m trying to write an algorithm for finding out the number of ways n numbers can be ordered. For example, two number say a and b can be ordered in 3 ways..

Similarly, 3 numbers can be arranged in 13 ways.

I found out that I can solve the problem using dynamic programming. And here’s is what I’m thinking to have layers which represent different ordering. Ex. a > b have two layers and a = b has a single layer and so on. So that I can use it for later purposes as done in dynamic programming. But I’m not able to write a recurrence relation for the same. Can someone suggest me how can I write that?

Answer 1

Assume f(n,k) = number of possible ways by having k inequality (and so n-k-1 equality), so:
assume you have n-1 number, now you want to add another number and calculate f(n,k), then we have two possibility:

1) There are (k-1) inequalities in those (n-1) numbers, and there are (k+1)(f(n-1,k-1)) ways to add n’th number so that new inequality added.

2) There are k inequalities in those (n-1) numbers, and there are (k+1)(f(n-1,k)) ways to add n’th number with no additional inequality.

f(n,k) = (k+1)(f(n-1,k-1) + f(n-1,k))

and what you want is sum of them (from zero to n-1), Bellow code is in c# (just tested for simple cases), in fact We just calculate number of possible ways not generating all ways..

class EqualInequalPermutation
{
    public int OrderingsNumber(int n)
    {
        int[][] f = new int[n+1][];
        for (int i = 0; i < n+1; i++)
        {
            f[i] = new int[n];
            for (int j = 0; j < n; j++)
                f[i][j] = 0;
        }
        f[1][0] = 1;
        int factorial = 1;
        for (int i = 1; i <= n; i++)
        {
            f[i][0] = 1;
            factorial *= i;
            f[i][i - 1] = factorial;
            for (int k = 1; k < n; k++)
            {
                f[i][k] = (k + 1) * (f[i - 1][k - 1] + f[i - 1][k]);
            }
        }
        int answer = 0;
        for (int i = 0; i < n; i++)
        {
            answer += f[n][i];
        }
        return answer;
    }
}