The Fisher-Yates shuffle is backward

Given a list of n elements, such as cards in a deck, what is the right way to shuffle the list? That is, what is the appropriate algorithm to (pseudo)randomly permute the elements in the list, so that each of the n! possible permutations is equally likely?

This is an interesting problem, in part because it is easy to get wrong. The standard, all-the-cool-kids-know-it response is the Fisher-Yates shuffle, consisting of a sequence of n-1 carefully specified random transpositions, with the following basic implementation in Python:

def fisher_yates_shuffle(a):
    """Shuffle list a[0..n-1] of n elements."""
    for i in range(len(a) - 1, 0, -1): # i from n-1 downto 1
        j = random.randint(0, i) # inclusive
        a[i], a[j] = a[j], a[i]

Note that the loop index i decreases from n-1 down to 1. Everywhere I have looked, this is how the algorithm is always presented. The motivation for this post is to wonder aloud why the following variant– which seems simpler, at least to me– is not the “standard” approach, with the only difference being that the loop runs “forward” instead of backward:

def forward_shuffle(a):
    "Shuffle list a[0..n-1] of n elements."""
    for i in range(1, len(a)): # i from 1 to n-1
        j = random.randint(0, i) # inclusive
        a[i], a[j] = a[j], a[i]

It’s worth emphasizing that this is different from what seems to be the usual “forward” version of the algorithm (e.g., this “equivalent version”), that seems to consistently insist on also “mirroring” the ranges of the random draws, so that they are decreasing with each loop iteration instead of the loop index:

def mirror_shuffle(a):
    "Shuffle list a[0..n-1] of n elements."""
    for i in range(0, len(a) - 1): # i from 0 to n-2
        j = random.randint(i, len(a) - 1) # inclusive
        a[i], a[j] = a[j], a[i]

There are a couple of ways to see and/or prove that forward_shuffle does indeed yield a uniform distribution on all possible permutations. One is by induction– the rather nice loop invariant is that, after each iteration i, the sublist a[0..i] is a uniformly random permutation of the original sublist a[0..i]. (Contrast this with the normal Fisher-Yates shuffle, where after each iteration indexed by i, the “suffix” sublist a[i..n-1] is essentially a uniformly permuted reservoir-sampled subset of the entire original list.)

Another way to see that forward_shuffle works as desired is to relate its behavior to that of the original fisher_yates_shuffle, which has already been proved correct. Consider the set of independent discrete random variables \{X_1, X_2, \ldots, X_{n-1}\}, with each X_i distributed uniformly between 0 and i, inclusive. These X_i are the random draws returned from random.randint(0, i).

Imagine generating the entire set of those n-1 independent random draws up front, then applying the sequence of corresponding transpositions (i, X_i). The original Fisher-Yates shuffle applies those transpositions in order of decreasing i, while forward_shuffle applies the same set of random transpositions, but in reverse order. Thus, the permutations resulting from fisher_yates_shuffle and forward_shuffle are inverses of each other… and if a random permutation is uniformly distributed, then so is its inverse.

There is nothing special here– indeed, this forward_shuffle is really just a less dressed-up implementation of what is usually referred to as the “inside-out” version of Fisher-Yates, that for some reason seems to be presented as only appropriate when shuffling a list generated from an external source (possibly of unknown length):

def forward_shuffle(source):
    "Return shuffled list from external source."""
    a = []
    for i, x in enumerate(source):
        j = random.randint(0, i) # inclusive
        a[i], a[j] = a[j], a[i]
    return a

I say “less dressed-up” because I’ve skipped what seems to be the usual special case j==i comparison that would eliminate the swapping. The above seems simpler to me, and I would be curious to know if these (branchless) swaps are really less efficient in practice.

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3 Responses to The Fisher-Yates shuffle is backward

  1. Provocative title! I’ve always written it the usual backwards way, and I hadn’t considered doing it forwards except by way of that clunky “equivalent version” that draws numbers from a range not starting at zero. That’s also an insightful observation about reservoir sampling. That never occurred to me.

    The Go, CPython, and OpenJSK all implement it the backwards way:

    I can’t think of any reason not to do it forwards, so maybe that’s how I’ll do it from now on.

  2. One thing is that in the “backwards” way, the hot zone, the zone that can be changed, gets smaller and smaller. And the “finished” zone gets larger and larger. Meanwhile in the “forwards” way, the hot zone gets larger and larger, and the finished zone does not exist until the very end.

    Maybe that makes a difference for keeping elements in memory or cache. You don’t have to potentially reload elements in the finished zone.

  3. David says:

    Along the lines of what Rohan said, with the usual implementation of Fisher Yates you don’t have to iterate over all of the items if you don’t need them all. Each iteration gives you a shuffled item.

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