Question

I’m reading about permutations and I’m interested in ranking/unranking methods.

From the abstract of a paper:

A ranking function for the permutations on n symbols assigns a unique
integer in the range [0, n! – 1] to each of the n! permutations. The corresponding
unranking function is the inverse: given an integer between 0 and n! – 1, the
value of the function is the permutation having this rank.

I made a ranking and an unranking function in C++ using next_permutation. But this isn’t practical for n>8. I’m looking for a faster method and factoradics seem to be quite popular.
But I’m not sure if this also works with duplicates. So what would be a good way to rank/unrank permutations with duplicates?

Answer 1

One way is to rank and unrank the choice of indices by a particular group of equal numbers, e.g.,

def choose(n, k):
    c = 1
    for f in xrange(1, k + 1):
        c = (c * (n - f + 1)) // f
    return c

def rank_choice(S):
    k = len(S)
    r = 0
    j = k - 1
    for n in S:
        for i in xrange(j, n):
            r += choose(i, j)
        j -= 1
    return r

def unrank_choice(k, r):
    S = []
    for j in xrange(k - 1, -1, -1):
        n = j
        while r >= choose(n, j):
            r -= choose(n, j)
            n += 1
        S.append(n)
    return S

def rank_perm(P):
    P = list(P)
    r = 0
    for n in xrange(max(P), -1, -1):
        S = []
        for i, p in enumerate(P):
            if p == n:
                S.append(i)
        S.reverse()
        for i in S:
            del P[i]
        r *= choose(len(P) + len(S), len(S))
        r += rank_choice(S)
    return r

def unrank_perm(M, r):
    P = []
    for n, m in enumerate(M):
        S = unrank_choice(m, r % choose(len(P) + m, m))
        r //= choose(len(P) + m, m)
        S.reverse()
        for i in S:
            P.insert(i, n)
    return tuple(P)

if __name__ == '__main__':
    for i in xrange(60):
        print rank_perm(unrank_perm([2, 3, 1], i))