Fixing Random, part 8

[Code is here.]

Last time on FAIC we sketched out an O(log n) algorithm for sampling from a weighted distribution of integers by implementing a discrete variation on the “inverse transform” method where the “inverse” step involves a binary search.

Can we do better than O(log n)? Yes — but we’ll need an entirely different method. I’ll sketch out two methods, and implement one this time, and the other in the next exciting episode.

Again, let’s suppose we have some weights, say 10, 11, 5. We have three possible results, and the highest weight is 11. Let’s construct a 3 by 11 rectangle that looks like our ideal histogram; I’ll put dashes in to more clearly indicate the “spaces”:


Here’s an algorithm for sampling from this distribution:

  • Uniformly choose a random row and column in the rectangle; it’s easy to choose a number from 0 to 2 for the row and a number from 0 to 10 for the column.
  • If that point is a *, then the sample is the generated row number.
  • If the point is a , try again.

Basically, we’re throwing darts at the rectangle, and the likelihood of hitting a valid point on particular row is proportional to the probability of that row.

In case that’s not clear, let’s work out the probabilities. To throw our dart, first we’ll pick a row, uniformly, and then we’ll pick a point in that row, uniformly. We have a 1/3 * 10/11 = 10/33 chance of hitting a star from row 0, an  1/3 * 11/11 = 11/33 chance of hitting a star from row 1, a 1/3 * 5/11 = 5/33 chance of hitting a star from row 2, and that leaves a 7/33 chance of going again. We will eventually pick a *, and the values generated will conform to the desired distribution.

Let’s implement it. We’ll throw away our previous attempt and start over. (No pun intended.)

private readonly List<IDistribution<int>> distributions;
private WeightedInteger(List<int> weights)
  this.weights = weights;
  this.distributions =
    new List<IDistribution<int>>(weights.Count);
  int max = weights.Max();
  foreach(int w in weights)
    distributions.Add(Bernoulli.Distribution(w, max  w));

All right, we have three distributions; in each, a zero is success and a one is failure. In our example of weights 10, 11, 5, the first distribution is “10 to 1 odds of success”, the second is “always success”, and the third is “5 to 6 odds of success”. And now we sample in a loop until we succeed. We uniformly choose a distribution, and then sample from that distribution until we get success.

public int Sample() 
  var rows = SDU.Distribution(0, weights.Count  1);
  while (true)
    int row = rows.Sample();
    if (distributions[row].Sample() == 0)
      return row;

We do two samples per loop iteration; how many iterations are we likely to do? In our example we’ll get a result on the first iteration 26/33 of the time, because we have 26 hits out of 33 possibilities. That’s 79% of the time. We get a result after one or two iterations 95% of the time. The vast majority of the time we are going to get a result in just a handful of iterations.

But now let’s think again about pathological cases. Remember our distribution from last time that had 1000 weights: 1, 1, 1, 1, …, 1, 1001. Consider what that histogram looks like.

   0|*------...---  (1 star, 1000 dashes)
   1|*------...---  (1 star, 1000 dashes)
 998|*------...---  (1 star, 1000 dashes)
 999|*******...***  (1001 stars, 0 dashes)

Our first example has 26 stars and 6 dashes. In our pathological example there will be 2000 stars and 999000 dashes, so the probability of exiting the loop on any particular iteration is about one in 500. We are typically going to loop hundreds of times in this scenario! This is far worse than our O(log n) option in pathological cases.

This algorithm is called “rejection sampling” (because we “reject” the samples that do not hit a *) and it works very well if all the weights are close to the maximum weight, but it works extremely poorly if there is a small number of high-weight outputs and a large number of low-weight outputs.

Fortunately there is a way to fix that problem. I’m going to reject rejection sampling, and move on to our third and final technique.

Let’s re-draw our original histogram, but I’m going to make two changes. First, instead of stars I’m going to fill in the number sampled, and I’m going to make it a 33 by 3 rectangle, and triple the size of every row.


Plainly this is logically no different; we could “throw a dart”, and the number that we hit is the sample; if we hit a dash, we go again. But we still have the problem that 21 out of 99 times we’re going to hit a dash.

My goal is to get rid of all the dashes, but I’m going to start by trying to get 7 dashes in each row. There are 21 dashes available, three rows, so that’s seven in each row.

To achieve that, I’m going to first pick an “excessive” row (too many numbers, too few dashes) and “deficient row” (too few numbers, too many dashes) and move some of the numbers from the excessive row to the deficient row, such that the deficient row now has exactly seven dashes. For example, I’ll move eleven of the 0s into the 2 row, and swap eleven of the 2 row’s dashes into the 0 row.


We’ve achieved our goal for the 2 row. Now we do the same thing again. I’m going to move seven of the 1s into the 0 row:


We’ve achieved our goal. And now we can get rid of the dashes without changing the distribution:


Now we can throw darts and never get any rejections!

The result is a graph where we can again, pick a column, and then from that column sample which result we want; there are only ever one or two possibilities in a column, so each column can be represented by a Bernoulli or Singleton distribution. Therefore we can sample from this distribution by doing two samples: a uniform integer to pick the row, and the second one from the distribution associated with that row.

Aside: You might wonder why I chose to move eleven 0s and then seven 1s. I didn’t have to! I also could have moved eleven 1s into the 2 row, and then four 0s into the 1 row. Doesn’t matter; either would work. As we’ll see in the next episode, as long as you make progress towards a solution, you’ll find a solution.

This algorithm is called the “alias method” because some portion of each row is “aliased” to another row. It’s maybe not entirely clear how to implement this algorithm but it’s quite straightforward if you are careful.

Next time on FAIC: The alias method implementation.


14 thoughts on “Fixing Random, part 8

  1. Do you still need rows and columns in your “Alias method” example? Shouldn’t you just be able to put the numbers into a single dimensioned array and sample it once?

    • That’s exactly the method we’re trying to avoid. The “build an array of outcomes” technique consumes O(total of all weights) memory and takes O(1) time, and that’s just too much memory for the time win. As we’ll see, the alias method consumes O(number of weights) memory and takes O(1) time, so it is a clear improvement.

  2. A pathological case for the third method would look like this I presume:


    Where if you sample the 7th element of the 0th row, finding it would be log(n) in the length of the row at best.

    • I think the point he’s getting to is that there are no pathological (sampling) cases for the alias method. If I am reading this correctly, he is saying that any non-uniform discrete distribution with n weights can be modeled as n Bernouli or singleton distributions. Generating the sub-distributions when you construction of the “outer” distribution may be difficult (which is possibly why he left the implementation for a separate post), but the whole idea is that you aren’t “finding elements in rows” anymore.

      So, for your proposed distribution, one possible set of bernouli distributions would be:

      B: 0:10, 1:81
      B: 1:19, 2:72
      B: 2:28, 3:63
      B: 3:37, 4:54
      B: 4:46, 5:45
      B: 5:55, 6:36
      B: 6:64, 7:27
      B: 7:73, 8:18
      B: 8:82, 9:9
      S: 9

      (I don’t feel like typing out 100 columns to prove it)

      So, for sampling we’re no longer O(n) or O(log n) but instead O(1) always.

      Constructing the distribution seems like it could be a pain in the butt though.

      • Having read the Darts, Dice, and Coins linked above, the algorithm doesn’t look that bad. Doing everything in weights and odds instead of probabilities seems to have made many things easier. I’m interested to see how that affects the algorithm.

        • Indeed; my pain point when implementing the algorithm originally was (you guessed it) floating point error in Go.

          Using integer weights suddenly makes that disappear, without requiring an arbitrary-precision Decimal.

  3. Pingback: Dew Drop – February 27, 2019 (#2908) | Morning Dew

  4. Pingback: Fixing Random, part 9 | Fabulous adventures in coding

  5. How do you go about discovering these various algorithms? They would certainly not occur to me if I were just presented with the problem to solve. What are good resources for going from a problem description to suitable algorithms to apply to the problem?

    • Good question. At the end of this series I’ll give a partial list of some of the papers and web sites I read while researching it.

      But your question is much more general; basically your question is “how do I solve problems in computer programming?” That is a HUGE topic, but a few ideas:

      1) Get a solid grounding in the fundamentals of data structures and algorithms. “Introduction To Algorithms” by Cormen, Leieserson, Rivest and Stein is the standard text. Knuth’s Art of Computer Programming is of course the classic, but not super approachable by the novice. Chris Okasaki’s book is my go-to source for analysis of immutable data structures.

      2) If those are too advanced for your current level, there are many modern introductory books on algorithmic thinking. Just do a web search, you’ll find plenty.

      3) Suppose your problem isn’t solvable by an application of basic sorting/searching data structures. Do a search for papers by researchers in that area, and see what they’ve tried. This is a great way to understand what the limits of current research are, and whether your problem is even solvable. All the time I see on StackOverflow people asking questions that are some variation on “how do I solve the knapsack problem quickly?” You don’t! And if you need a partial solution, consult the literature.

      4) A technique I often use is to suppose that I have an “oracle” — that is, a device that can solve the problem efficiently. I write a class with a method whose signature has exactly the inputs and outputs I need. I can then put in enough gear in the implementation of the method to get the program compiling and passing one test case, even if I don’t know how to solve the problem in general. I can then fill in different implementations and try them out against my test cases.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s