Reply To: Programming

Looking at it, I realize that I called factorial() twice for each iteration, which is too inefficient. Here, I fixed it:

def nth_permutation(n,objects):
    perm = []
    n -= 1
    while len(objects) > 0:
        k = factorial(len(objects)-1)
        m = int(n/k)
        n -= m * k
        perm.append(objects[m])
        objects.pop(m)
    return perm

@ByteInBits, you said

Normally I would smile and let it pass but I have to take issue with you
to come to some understanding.

In post #98171 you state:
@ByteInBits, your MD5 sum is for the NEXT permutation. By my example, you should start with 1 and not 0.

This make you incorrect with the entry number!!
(standard practice doese not start at 1, and you never stated the deviation)

But I did. When I said:

Notice that I listed them in a sort of numerical order: 123, 132, 213, 231, 312, 321. There are 3! = 6 of them. If I were to ask you what is the fourth permutation of 1, 2, 3, you should answer “2, 3, 1”. OK, got it?

I was very careful about making sure the list started at 1, so I said “fourth” instead of “third”. I even stopped to ask if everyone understood before I went on.

Never wager against a Sicilian when death is on the line.