python - Generate a random derangement of a list

Question

How can I randomly shuffle a list so that none of the elements remains in its original position?

In other words, given a list A with distinct elements, I'd like to generate a permutation B of it so that

• this permutation is random
• and for each n, a[n] != b[n]

e.g.

a = [1,2,3,4]
b = [4,1,2,3] # good
b = [4,2,1,3] # good

a = [1,2,3,4]
x = [2,4,3,1] # bad

I don't know the proper term for such a permutation (is it "total"?) thus having a hard time googling. The correct term appears to be "derangement".

Answer 1

After some research I was able to implement the "early refusal" algorithm as described e.g. in this paper. It goes like this:

import random

def random_derangement(n):
    while True:
        v = [i for i in range(n)]
        for j in range(n - 1, -1, -1):
            p = random.randint(0, j)
            if v[p] == j:
                break
            else:
                v[j], v[p] = v[p], v[j]
        else:
            if v[0] != 0:
                return tuple(v)

The idea is: we keep shuffling the array, once we find that the permutation we're working on is not valid (v[i]==i), we break and start from scratch.

A quick test shows that this algorithm generates all derangements uniformly:

N = 4

# enumerate all derangements for testing
import itertools
counter = {}
for p in itertools.permutations(range(N)):
    if all(p[i] != i for i in p):
        counter[p] = 0

# make M probes for each derangement
M = 5000
for _ in range(M*len(counter)):
    # generate a random derangement
    p = random_derangement(N)
    # is it really?
    assert p in counter
    # ok, record it
    counter[p] += 1

# the distribution looks uniform
for p, c in sorted(counter.items()):
    print p, c

Results:

(1, 0, 3, 2) 4934
(1, 2, 3, 0) 4952
(1, 3, 0, 2) 4980
(2, 0, 3, 1) 5054
(2, 3, 0, 1) 5032
(2, 3, 1, 0) 5053
(3, 0, 1, 2) 4951
(3, 2, 0, 1) 5048
(3, 2, 1, 0) 4996

I choose this algorithm for simplicity, this presentation briefly outlines other ideas.