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sort_by{rand}'s bias makes BigDecimal cry

sort_by{ rand } is a common idiom to shuffle an array in Ruby. It's short, looks good and is fairly efficient (despite being O ( n roman log n )) since Array#sort_by is done at C speed.

Unfortunately, it is biased: if the same rand() value is returned for two different elements, their relative order will be preserved.

%w[a b c d].sort_by{ 0 }                           # => ["a", "b", "c", "d"]
i = 0
%w[a b c d].sort_by{ 10 - (i += 1) / 2 }           # => ["d", "b", "c", "a"]

This means that permutations preserving the relative order of one (or more) pair of elements of the original array are a bit more probable.

How often will that happen? The Mersenne Twister PRNG used by Ruby has been proven to have a period of 2 sup 19937 - 1, so it would seem at first sight that the answer is "not before the sun turns into a nova". But Kernel#rand does not return the full state of the PRNG: you only get 53 bits of pseudo-randomness out of a call to rand(). This is becoming more interesting.

The birthday paradox

The birthday paradox states that given 23 people in a room, the odds are over 50% that two of them will have the same birthday. It is not actually a paradox (there's no self-contradiction), it merely defies common intuition.

It turns out that the problem with sort_by{ rand } is another instance of the birthday paradox. In this case, we have a room with N people (N being the size of the array to shuffle) and there are 2 sup 53 "days in a year".

Let's try to reproduce the 23-50% result, and apply the method to the other one.

There are

roman A(365,m) = { 365! } over {left ( 365 - m right ) ! }

ways to assign a birthday to m people without any repetition. Therefore, the probability that at least two people share a birthday is

1 - {roman A(365,m)} over { 365 sup m }

In Ruby terms, we first have to define factorial:

class Integer
  def fact0
    case self
    when 0..1
      self * (self - 1).fact0

This is fairly slow and will typically fail for 6000! or so, depending on the stack size. Stirling's approximation proves helpful here: we'll use

n! approx sqrt{ 2 pi } n sup {n + 1 over 2} e sup{-n}

class Integer  
  def fact
    Math.sqrt(2*Math::PI*self) * Math.exp(-self) * Math.exp(self * Math.log(self))

It compares rather well to the recursive factorial:

%w[fact0 fact].each do |m|
  t = Time.new
  eval <<-EOF
    200.times{|i| (i+1).#{m}}
  "#{m}: #{Time.new - t}"     # => "fact0: 0.125991", "fact: 0.001527"

Time to define A(n,m):

def A(n,m)
  n.fact / (n-m).fact

def C(n,m)
  A(n,m) / m.fact

Given that definition, the well-known 23-50% figures should be easy to reproduce:

(21..23).each do |i|
  [i, "%5.3f" % (1 - A(365,i)/(365**i))] # => [21, "  nan"], [22, "  nan"], [23, "  nan"]

Something's wrong, and the culprit is not hard to locate:

A(365, 20)                                         # => NaN

BigDecimal to the rescue

Ruby's standard library includes an extension for arbitrary precision floats, BigDecimal, whose very name suggests that we can think of it as Bignum's cousin. BigDecimal objects have to be initialized with a String, but that's not much of a problem

require 'bigdecimal'
require 'bigdecimal/math'

class Integer
  BigMath = Class.new{ extend ::BigMath }
  def fact(prec = 10) # !> method redefined; discarding old fact
    BigDecimal.new((Math.sqrt(2*Math::PI*self) * Math.exp(-self)).to_s) *
    BigMath.exp(BigDecimal.new(self.to_s) * BigMath.log(BigDecimal.new(self.to_s), prec), prec)

Let's go back to the birthday problem:

(21..23).each do |i|
  [i, "%5.3f" % (1 - A(365,i)/(365**i))]  # => [21, "0.444"], [22, "0.476"], [23, "0.507"]


Quantifying the bias

I'd now like to just do something like

 A(2**53, 1000000) / (2**53) ** 1000000

but it's obvious that won't fly. This is where a tiny bit of mathematical insight proves much more useful than anything I could possibly find in the standard library. I can find a cheap lower bound for the probability of repetition as follows: there are roman C(N,2) = {N left ( N - 1 right )} over 2 possible pairs for the N picks of rand() I'll do. For each of them, the probability that the second has got the same value is 1 - 2 sup{-53}. So the probability of having at least one repetition is going to be higher than

{ left ( 1 - 2 sup {-53} right ) } sup { roman C(N,2) }

which can be approximated as

1 - roman C(N,2) 2 sup {-53}

Let's code that:

include BigMath
N = 2 ** 53
PICKS = [3, 4, 6, 7].map{|n| 10 ** n}

Math.log(N)/Math.log(10)                           # => 15.954589770191

PICKS.each do |i|
  picks = BigDecimal.new(i.to_s)
  p_bias = "%7.5e" % (picks * (picks - 1) / (2 * N))
  [i, p_bias]       # => [1000, "5.54556e-11"], [10000, "5.55056e-09"], 
                    # [1000000, "5.55111e-05"], [10000000, "5.55111e-03"]


 arr.sort_by{ rand }

is going to be biased for an array with 10 million elements with a probability of around 0.55%. Just to make sure those figures are OK, here's another way to estimate that probability:

PICKS.each do |i|
  picks = BigDecimal.new(i.to_s)
  p_bias = "%7.5e" % (1.0 - exp(-picks*(picks -1) / (2*N), 100))
  [i, p_bias]       # => [1000, "5.54558e-11"], [10000, "5.55056e-09"], 
                    #    [1000000, "5.55096e-05"], [10000000, "5.53574e-03"]

Back to sort_by

While there are better ways to shuffle an array in theory (both less complex and unbiased), I'll keep using sort_by{rand} for small arrays.

lee b. 2006-02-16 (Thr) 21:25:34

def recfact(hi, lo=2) (hi-lo>42) ? recfact(hi,(hi+lo)/2) * recfact((hi+lo)/2 - 1, lo) : (lo..hi).inject {|f,i| f*i} end

Last modified:2005/12/05 12:29:20
Keyword(s):[blog] [ruby] [sort_by] [rand] [bigdecimal] [birthday] [paradox]