Probability Distribution for Programmers
Join the DZone community and get the full member experience.
Join For FreeThis is one of these very simple programming puzzles I came across recently:
given a function returning random integers fromOf course the solution should also return equally distributed numbers. So let’s start from an input function sample definition:0
to4
inclusive with equal probability, write a function returning random integers from0
to6
inclusive.
def rand4() = (math.random * 5).toIntYour task is to implement
rand6()
by only using rand4()
. Give yourself few minutes and continue reading..
.
.
The first approach is pretty straightforward but happens to be completely broken:
def rand6() = rand4() * 3 / 2As simple as that. In ideal solution each output value from
0
to 6
should appear with the probability of 1/7
. Can you tell from the code above, what's the probability of rand6()
returning 2
or 5
? That's right, it's no more than 0
, you'll never get these values. I hope it's clear why. So let's go for something more sophisticated:def rand6() = (rand4() + rand4()) % 7Looks better, but still pretty far. The code above has two major flaws. First of all the results of
rand4() + rand4()
expression range from 0
to 8
but we need 0
to 6
. The obvious solution is to use % 7
operation. However this results in 0
and 1
being returned twice as often because 7
and 8
are overflowing to 0
and 1
. So what about this:def rand6(): Int = { val rand8 = rand4() + rand4() if(rand8 > 6) rand6() else rand8 }I hope the recursion (which can easily be turned into loop, but I leave that work to the Scala compiler) is not obscuring the intent  if the sum of two
rand4()
invocations is above expected result, we simply discard it and call rand6()
again. However there is still one subtle but striking bug, quoting Wikipedia on uniform distributionThe sum of two independent, equally distributed, uniform distributions yields a symmetric triangular distribution.If you don't quite get the above, have a look at this live demo in JavaScript using
<canvas/>
illustrating what Wikipedia means:This program simply places pixels at random
(X, Y)
positions on each panel. In the first panel I use one Math.random() * 300
call scaled to fit whole canvas. As you can see the distribution is
more or less uniform. But we can't tell that about second and third
panels. On the second panel I am using the sum of two uniformly
distributed variables, in principle: (Math.random() + Math.random()) * 150)
. Even though this expression can return anything between 0
and 300
,
the points are very biased toward the middle of the canvas (triangular
distribution). The same behaviour is emphasized on the third panel where
ten invocations of Math.random()
are used.The correct answer
The approach I'm taking is based on the observation thatrand4()
is capable of producing two random least significant bits. So let's start from implementing rand3()
with known semantics:def rand3(): Int = rand4() match { case 4 => rand3() case x => x }
rand3()
returns uniformly distributed values from 0
to 3
doing so by rejecting 4
output of rand4()
. How will that help us? Well, we now have two random bits, each one being either 0
or 1
with 50% probability. We can easily widen it for larger sequences, e.g. rand15()
and rand7()
:def rand15() = (rand3() << 2) + rand3() def rand7() = rand15() >> 1You should be rather comfortable with the bit fiddling above. Having the ability to produce two random bits I can easily generate 4 and 3. Now
rand6()
is a nobrainer:def rand6() = rand7() match { case 7 => rand6() case x => x }Just to make this lesson a little bit more interesting, let's implement
randN(n: Int)
on top of rand4()
. randN()
should return uniformly distributed natural values from 0
to n
. I'll begin by implementing helper method atLeastKrandBits(k: Int)
returning... at least K random bits:def atLeastKrandBits(k: Int): Int = k match { case 0 => 0 case 1 => rand3() >> 1 case 2 => rand3() case b => (atLeastKrandBits(k  2) << 2) + rand3() }Alternative implementation with
foldLeft()
: def atLeastKrandBits(k: Int) = (0 to (k + 1) / 2).foldLeft(0){ (acc, _) => (acc << 2) + rand3() }...or if you really hate those to maintain your code:
def atLeastKrandBits(k: Int) = (0 /: (0 to (k + 1) / 2)){ (acc, _) => (acc << 2) + rand3() }Having any of the implementations above
randN(n: Int)
is simple:def randN(n: Int) = { val bitsCount = java.lang.Integer.highestOneBit(n) val randBits = atLeastKrandBits(bitsCount) if(randBits > n) randN(n) else randBits }
Conclusions
You might ask yourself a question: why should I even care? If you fail to understand probability distribution your application might produce random output that's actually quite easy to predict. It's not a big deal if you are writing a game and enemies are more likely to appear at some places on the map (but the players will discover and abuse it!) But if you need random numbers for security reasons or you rely on uniform distribution e.g. for loadbalancing purposes  any bias might become fatal.
Distribution (differential geometry)
Published at DZone with permission of Tomasz Nurkiewicz, DZone MVB. See the original article here.
Opinions expressed by DZone contributors are their own.
Trending

Why You Should Consider Using React Router V6: An Overview of Changes

MLOps: Definition, Importance, and Implementation

Mastering Time Series Analysis: Techniques, Models, and Strategies

Playwright JavaScript Tutorial: A Complete Guide
Comments