Introducing Bean Machine

The final part of my Life series is still in the works but I need to interrupt that series with some exciting news. The new programming language I have been working on for the last year or so has just been announced by the publication of our paper Bean Machine: A Declarative Probabilistic Programming Language For Efficient Programmable Inference


Before I get into the details, a few notes on attributing credit where it is due and the like:

  • Though my name appears on the paper as a courtesy, I did not write this paper. Thanks and congratulations in particular to Naz Tehrani and Nim Arora who did a huge amount of work getting this paper together.
  • The actual piece of the language infrastructure that I work on every day is a research project involving extraction, type analysis and optimization of the Bayesian network underlying a Bean Machine program. We have not yet announced the details of that project, but I hope to be able to discuss it here soon.
  • Right now we’ve only got the paper; more information about the language and how to take it out for a spin yourself will come later. It will ship when its ready, and that’s all the scheduling information I’ve got.
  • The name of the language comes from a physical device for visualizing probability distributions because that’s what it does.


I will likely do a whole series on Bean Machine later on this autumn, but for today let me just give you the brief overview should you not want to go through the paper. As the paper’s title says, Bean Machine is a Probabilistic Programming Language (PPL).

For a detailed introduction to PPLs you should read my “Fixing Random” series, where I show how we could greatly improve support for analysis of randomness in .NET by both adding types to the base class library and by adding language features to a language like C#.

If you don’t want to read that 40+ post introduction, here’s the TLDR.

We are all used to two basic kinds of programming: produce an effect and compute a result. The important thing to understand is that Bean Machine is firmly in the “compute a result” camp. In our PPL the goal of the programmer is to declaratively describe a model of how the world works, then input some observations of the real world in the context of the model, and have the program produce posterior distributions of what the real world is probably like, given those observations. It is a language for writing statistical model simulations.

A “hello world” example will probably help. Let’s revisit a scenario I first discussed in part 30 of Fixing Random: flipping a coin that comes from an unfair mint. That is, when you flip a coin from this mint, you do not necessarily have a 50-50 chance of getting heads vs tails. However, we do know that when we mint a coin, the distribution of fairness looks like this:

Fairness is along the x axis; 0.0 means “always tails”, 1.0 means “always heads”. The probability of getting a coin of a particular fairness is proportional to the area under the graph. In the graph above I highlighted the area between 0.6 and 0.8; the blue area is about 25% of the total area under the curve, so we have a 25% chance that a coin will be between 0.6 and 0.8 fair.

Similarly, the area between 0.4 and 0.6 is about 30% of the total area, so we have a 30% chance of getting a coin whose fairness is between 0.4 and 0.6. You see how this goes I’m sure.

Suppose we mint a coin; we do not know its true fairness, just the distribution of fairness above. We flip the coin 100 times, and we get 72 heads, 28 tails. What is the most probable fairness of the coin?

Well, obviously the most probable fairness of a coin that comes up heads 72 times out of 100 is 0.72, right?

Well, no, not necessarily right. Why? Because the prior probability that we got a coin that is between 0.0 and 0.6 is rather a lot higher than the prior probability that we got a coin between 0.6 and 1.0. It is possible by sheer luck to get 72 heads out of 100 with a coin between 0.0 and 0.6 fairness, and those coins are more likely overall.


Aside: If that is not clear, try thinking about an easier problem that I discussed in my earlier series. You have 999 fair coins and one double-headed coin. You pick a coin at random, flip it ten times and get ten heads in a row. What is the most likely fairness, 0.5 or 1.0? Put another way: what is the probability that you got the double-headed coin? Obviously it is not 0.1%, the prior, but nor is it 100%; you could have gotten ten heads in a row just by luck with a fair coin. What is the true posterior probability of having chosen the double-headed coin given these observations?


What we have to do here is balance between two competing facts. First, the fact that we’ve observed some coin flips that are most consistent with 0.72 fairness, and second, the fact that the coin could easily have a smaller (or larger!) fairness and we just got 72 heads by luck. The math to do that balancing act to work out the true distribution of possible fairness is by no means obvious.

What we want to do is use a PPL like Bean Machine to answer this question for us, so let’s build a model!

The code will probably look very familiar, and that’s because Bean Machine is a declarative language based on Python; all Bean Machine programs are also legal Python programs. We begin by saying what our “random variables” are.


Aside: Statisticians use “variable” in a way very different than computer programmers, so do not be fooled here by your intuition. By “random variable” we mean that we have a distribution of possible random values; a representation of any single one of those values drawn from a distribution is a “random variable”. 


To represent random variables we declare a function that returns a pytorch distribution object for the distribution from which the random variable has been drawn. The curve above is represented by the function beta(2, 2), and we have a constructor for an object that represents that distribution in the pytorch library that we’re using, so:

@random_variable
def coin():
  return Beta(2.0, 2.0)

Easy as that. Every usage in the program of coin() is logically a single random variable; that random variable is a coin fairness that was generated by sampling it from the beta(2, 2) distribution graphed above.


Aside: The code might seem a little weird, but remember we do these sorts of shenanigans all the time in C#. In C# we might have a method that looks like it returns an int, but the return type is Task<int>; we might have a method that yield returns a double, but the return type is IEnumerable<double>. This is very similar; the method looks like it is returning a distribution of fairnesses, but logically we treat it like a specific fairness drawn from that distribution.


What do we then do? We flip a coin 100 times. We therefore need a random variable for each of those coin flips:

@random_variable
def flip(i):
  return Bernoulli(coin())

Let’s break that down. Each call flip(0), flip(1), and so on on, are distinct random variables; they are outcomes of a Bernoulli process — the “flip a coin” process — where the fairness of the coin is given by the single random variable coin(). But every call to flip(0) is logically the same specific coin flip, no matter how many times it appears in the program.

For the purposes of this exercise I generated a coin and simulated 100 coin tosses to simulate our observations of the real world. I got 72 heads. Because I can peek behind the curtain for the purposes of this test, I can tell you that the coin’s true fairness was 0.75, but of course in a real-world scenario we would not know that. (And of course it is perfectly plausible to get 72 heads on 100 coin flips with a 0.75 fair coin.)

We need to say what our observations are.  The Bernoulli distribution in pytorch produces a 1.0 tensor for “heads” and a 0.0 tensor for “tails”. Our observations are represented as a dictionary mapping from random variables to observed values.

heads = tensor(1.0)
tails = tensor(0.0)
observations = {
  flip(0) : heads,
  flip(1) : tails,
  ...  and so on, 100 times with 72 heads, 28 tails.
}

Finally, we have to tell Bean Machine what to infer. We want to know the posterior probability of fairness of the coin, so we make a list of the random variables we care to infer posteriors on; there is only one in this case.

inferences = [ coin() ]
posteriors = infer(observations, inferences)
fairness = posteriors[coin()]

and we get an object representing samples from the posterior fairness of the coin given these observations. (I’ve simplified the call site to the inference method slightly here for clarity; it takes more arguments to control the details of the inference process.)

The “fairness” object that is handed back is the result of efficiently simulating the possible worlds that get you to the observed heads and tails; we then have methods that allow you to graph the results of those simulations using standard graphing packages:

The orange marker is our original guess of observed fairness: 0.72. The red marker is the actual fairness of the coin used to generate the observations, 0.75. The blue histogram shows the results of 1000 simulations; the vast majority of simulations that produced those 72 heads had a fairness between 0.6 and 0.8, even though only 25% of the coins produced by the mint are in that range.  As we would hope, both the orange and red markers are near the peak of the histogram.

So yes, 0.72 is close to the most likely fairness, but we also see here that a great many other fairnesses are possible, and moreover, we clearly see how likely they are compared to 0.72. For example, 0.65 is also pretty likely, and it is much more likely than, say, 0.85. This should make sense, since the prior distribution was that fairnesses closer to 0.5 are more likely than those farther away; there’s more “bulk” to the histogram to the left than the right: that is the influence of the prior on the posterior!

Of course because we only did 1000 simulations there is some noise; if we did more simulations we would get a smoother result and a clear, single peak. But this is a pretty good estimate for a Python program with six lines of model code that only takes a few seconds to run.


Why do we care about coin flips? Obviously we don’t care about solving coin flip problems for their own sake. Rather, there are a huge number of real-world problems that can be modeled as coin flips where the “mint” produces unfair coins and we know the distribution of coins that come from that mint:

  • A factory produces routers that have some “reliability”; each packet that passes through each router in a network “flips a coin” with that reliability; heads, the packet gets delivered correctly, tails it does not. Given some observations from a real data center, which is the router that is most likely to be the broken one? I described this model in my Fixing Random series.
  • A human reviewer classifies photos as either “a funny cat picture” or “not a funny cat picture”. We have a source of photos — our “mint” — that produces pictures with some probability of them being a funny cat photo, and we have human reviewers each with some individual probability of making a mistake in classification. Given a photo and ten classifications from ten reviewers, what is the probability that it is a funny cat photo? Again, each of these actions can be modeled as a coin flip.
  • A new user is either a real person or a hostile robot, with some probability. The new user sends a friend request to you; you either accept it or reject it based on your personal likelihood of accepting friend requests. Each one of these actions can be modeled as a coin flip; given some observations of all those “flips”, what is the posterior probability that the account is a hostile robot?

And so on; there are a huge number of real-world problems we can solve just with modeling coin flips, and Bean Machine does a lot more than just coin flip models!


I know that was rather a lot to absorb, but it is not every day you get a whole new programming language to explain! In future episodes I’ll talk more about how Bean Machine works behind the scenes, how we traded off between declarative and imperative style, and that sort of thing. It’s been a fascinating journey so far and I can’t hardly wait to share it.

 

Life, part 35

Last time we implemented what looked like Gosper’s algorithm and got a disappointing result; though the quadtree data structure is elegant and the recursive algorithm is simple, and even though we memoize every operation, the time performance is on par with our original naive implementation, and the amount of space consumed by the memoizers is ginormous. But as I said last time, I missed a trick in my description of the algorithm, and that trick is the key to the whole thing. (Code for this episode is here.)

One reader pointed out that we could be doing a better job with the caching. Sure, that is absolutely true. There are lots of ways we could come up with a better cache mechanism than my hastily-constructed dictionary, and those would in fact lead to marginal performance gains. But I was looking for a win in the algorithm itself, not in the details of the cache.

A few readers made the astute observation that the number of recursions — nine — was higher than necessary. The algorithm I gave was:

  • We are given an n-quad and wish to step the center (n-1)-quad.
  • We make nine unstepped (n-1)-quads and step each of them to get nine stepped (n-2)-quads
  • We reform those nine (n-2)-quads into four stepped (n-1)-quads, take the centers of each, and that’s our stepped (n-1) quad.

But we have all the information we need in the original n-quad to extract four unstepped (n-1)-quads. We then could step each of those to get four center stepped (n-2)-quads, and we can reform those into the desired (n-1)-quad.

Extracting those four unstepped (n-1)-quads is a fair amount of work, but there is an argument to be made that it might be worth the extra work in order to go from nine recursions to four. I didn’t try it, but a reader did and reports back that it turns out this is not a performance win. Regardless though, this wasn’t the win I was looking for.

Let’s go through the derivation one more time, and derive Gosper’s algorithm for real.

We still have our base case: we can take any 2-quad and get the center 1-quad stepped one tick forward. Suppose once again we are trying to step the outer green 3-quad forward; we step each of its component green 2-quads forward one tick to get these four blue 1-quads:

We then extract the north, south, east, west and center 2-quads from the 3-quad and step each of those forwards one tick, and that gives us these nine blue 1-quads, each one step in the future:

 

We then form four 2-quads from those nine 1-quads; here we are looking at the northwest 2-quad and its center 1-quad:

The light blue 2-quad and its dark blue 1-quad center are both one tick ahead of the outer green 3-quad. This is where we missed our trick.

We have the light blue 2-quad, and it is one tick ahead of the green 3-quad. We want to get its center 1-quad. What if we got its center 1-quad stepped one tick ahead? We know we can do it! It’s a 2-quad and we can get the center 1-quad of any 2-quad stepped one tick ahead. We can make the innermost dark blue quad stepped two ticks ahead. We repeat that operation four times and we have enough information to construct…

…the center 2-quad stepped two ticks ahead, not one.

Now let’s do the same reasoning for a 4-quad.

We step its nine component 3-quads forwards two ticks, because as we just saw, we can do that for a 3-quad. We then compose those nine 2-quads into four 3-quads, step each of those forward two ticks, again because we can, and construct the center 3-quad stepped four ticks ahead.

And now let’s do the same reasoning for an n-quad… you see where this is going I’m sure.

This is the astonishing power of Gosper’s algorithm. Given an n-quad, we can step forward its center (n-1)-quad by 2n-2 ticks for any n>=2.

Want to know the state of the board a million ticks in the future? Embiggen the board until it is a 22-quad — we know that operation is cheap and easy — and you can get the center 21-quad stepped forwards by 220 ticks using this algorithm. A billion ticks? Embiggen it to a 32-quad, step it forward 230 ticks.

We showed last time an algorithm for stepping an n-quad forward by one tick; here we’ve sketched an algorithm for stepping an n-quad forward by 2n-2 ticks. What would be really nice from a user-interface perspective is if we had a hybrid algorithm that can step an n-quad forward by 2k ticks for any k between 0 and n-2.

You may recall that many episodes ago I added an exponential “speed factor” where the factor is the log2 of the number of ticks to step. We can now write an implementation of Gosper’s algorithm for real this time that takes a speed factor. Rather than try to explain it further, let’s just look at the code.

private static Quad UnmemoizedStep((Quad q, int speed) args)
{
  Quad q = args.q;
  int speed = args.speed;

  Debug.Assert(q.Level >= 2);
  Debug.Assert(speed >= 0);
  Debug.Assert(speed <= q.Level - 2);

  Quad r;
  if (q.IsEmpty)
    r = Quad.Empty(q.Level - 1);
  else if (speed == 0 && q.Level == 2)
    r = StepBaseCase(q);
  else
  {
    // The recursion requires that the new speed be not
    // greater than the new level minus two. Decrease speed
    // only if necessary.
    int nineSpeed = (speed == q.Level - 2) ? speed - 1 : speed;
    Quad q9nw = Step(q.NW, nineSpeed);
    Quad q9n = Step(q.N, nineSpeed);
    Quad q9ne = Step(q.NE, nineSpeed);
    Quad q9w = Step(q.W, nineSpeed);
    Quad q9c = Step(q.Center, nineSpeed);
    Quad q9e = Step(q.E, nineSpeed);
    Quad q9sw = Step(q.SW, nineSpeed);
    Quad q9s = Step(q.S, nineSpeed);
    Quad q9se = Step(q.SE, nineSpeed);
    Quad q4nw = Make(q9nw, q9n, q9c, q9w);
    Quad q4ne = Make(q9n, q9ne, q9e, q9c);
    Quad q4se = Make(q9c, q9e, q9se, q9s);
    Quad q4sw = Make(q9w, q9c, q9s, q9sw);

    // If we are asked to step forwards at speed (level - 2), 
    // then we know that the four quads we just made are stepped 
    // forwards at (level - 3). If we step each of those forwards at 
    // (level - 3) also, then we have the center stepped forward at 
    // (level - 2), as desired.
    //
    // If we are asked to step forwards at less than speed (level - 2)
    // then we know the four quads we just made are already stepped
    // that amount, so just take their centers.

    if (speed == q.Level - 2)
    {  
      Quad rnw = Step(q4nw, speed - 1);
      Quad rne = Step(q4ne, speed - 1);
      Quad rse = Step(q4se, speed - 1);
      Quad rsw = Step(q4sw, speed - 1);
      r = Make(rnw, rne, rse, rsw);
    }
    else
    {
      Quad rnw = q4nw.Center;
      Quad rne = q4ne.Center;
      Quad rse = q4se.Center;
      Quad rsw = q4sw.Center;
      r = Make(rnw, rne, rse, rsw);
    }
  }
  Debug.Assert(q.Level == r.Level + 1);
  return r;
}

As I’m sure you’ve guessed, yes, we’re going to memoize this too! This power has not come for free; we are now doing worst case 13 recursions per non-base call, which means that we could be doing worst case 13n-3 base case calls in order to step forwards 2n-2 ticks, and that’s a lot of base case calls. How on earth is this ever going to work?

Again, because (1) we are automatically skipping empty space of every size; if we have an empty 10-quad that we’re trying to step forwards 256 ticks, we immediately return an empty 9-quad, and (2) thanks to memoization every time we encounter a problem we’ve encountered before, we just hand back the solution. The nature of Life is that you frequently encounter portions of boards that you’ve seen before because most of a board is stable most of the time. We hope.

That’s the core of Gosper’s algorithm, finally. (Sorry it took 35 episodes to get there, but it was a fun journey!) Let’s now integrate that into our existing infrastructure; I’ll omit the memoization and cache management because it’s pretty much the same as we’ve seen already.

The first thing to note is that we can finally get rid of this little loop:

public void Step(int speed)
{
  for (int i = 0; i < 1L << speed; i += 1)
    Step();
}

Rather than implementing Step(speed) in terms of Step(), we’ll go the other way:

public void Step()
{
  Step(0);
}

public void Step(int speed)
{
  // Cache management omitted
  const int MaxSpeed = MaxLevel - 2;
  Debug.Assert(speed >= 0);
  Debug.Assert(speed <= MaxSpeed);

The embiggening logic needs to be a little more aggressive. This implementation is probably more aggressive than we need it to be, but remember, empty space is essentially free both in space and processing time.

  Quad current = cells;
  if (!current.HasAllEmptyEdges)
    current = current.Embiggen().Embiggen();
  else if (!current.Center.HasAllEmptyEdges)
    current = current.Embiggen();
  while (current.Level < speed + 2)
    current = current.Embiggen();

  Quad next = Step(current, speed);
  cells = next.Embiggen();
  generation += 1L << speed;
  // Cache reset logic omitted
}

Now how are we going to perf test this thing? We already know that calculating 5000 individual generations of “acorn” with Gosper’s algorithm will be as slow as the original naïve version. What happens if for our performance test we set up acorn and then call Step(13)? That will step it forwards 8196 ticks:

Algorithm           time(ms) size  Mcells/s 
Naïve (Optimized):   4000     8      82     
Abrash (Original)     550     8     596     
Stafford              180     8    1820     
QuickLife              65    20      ?      
Gosper, sp 0 * 5000  3700    60      ?
Gosper, sp 13 * 1     820    60      ?

Better, but still not as good as any of our improvements over the naïve algorithm, and 13x slower than QuickLife.

So this is all very interesting, but what’s the big deal?

Do you remember the asymptotic time performance of Hensel’s QuickLife? It was O(changes); that is, the cost of computing one tick forwards is proportional to the number of changed cells on that tick. Moreover, period-two oscillators were essentially seen as not changing, which is a huge win.

We know that the long-term behaviour of acorn is that shortly after 5000 ticks in, we have only a handful of gliders going off to infinity and all the rest of the living cells are either still Lifes or period-two oscillators that from QuickLife’s perspective, might as well be still Lifes. So in the long run, the only changes that QuickLife has to process are the few dozens of cells changed for each glider; everything else gets moved into the “stable” bucket.

Since in the long run QuickLife is processing the same number of changes per tick, we would expect that the total time taken to run n ticks of acorn with QuickLife should grow linearly. Let’s actually try it out to make sure. I’m going to run one ticks of acorn with QuickLife, then reset, then run two ticks , then reset, then run four ticks, reset, eight ticks, and so on, measuring the time for each, up to 221 =~ 2.1 million ticks.

Here is a graph of the results; milliseconds on the y axis, ticks on the x axis, log-log scale. Lower is faster.


Obviously the leftmost portion of the graph is wrong; anything less than 256 ticks takes less than 1 millisecond but I haven’t instrumented my implementation to measure sub-millisecond timings because I don’t care about those. I’ve just marked all of them as taking one millisecond.

Once we’re over a millisecond, you can see that QuickLife’s time to compute some number of ticks grows linearly; it’s about 8 microseconds per tick, which is pretty good. You can also see that the line changes slope slightly once we get to the point where it is only the gliders on the active list; the slope gets shallower, indicating that we’re taking less time for each tick.

Now let’s do the same with Gosper’s algorithm; of course we will make sure to reset the caches between every run! Otherwise we would be unfairly crediting speed improvements in later runs to cached work that was done in earlier runs.

Hensel’s QuickLife in blue, Gosper’s HashLife in orange:

Holy goodness! 

The left hand side of the graph shows that Gosper’s algorithm is consistently around 16x slower than QuickLife in the “chaos” part of acorn’s evolution, right up to the point where we end up in the “steady state” of just still Lifes, period-two oscillators and gliders. The right hand side of the graph shows that once we are past that point, Gosper’s algorithm becomes O(1), not O(changes).

In fact this trend continues. We can compute a million, a billion, a trillion, a quadrillion ticks of acorn in around 800ms. And we can embiggen the board to accurately track the positions of those gliders even when they are a quadrillion cells away from the center.

What is the takeaway here? The whole point of this series is: you can take advantage of characteristics of your problem space to drive performance improvements. But what we’ve just dramatically seen here is that this maxim is not sufficient. You’ve also got to think about specific problems that you are solving.

Let’s compare and contrast. Hensel’s QuickLife algorithm excels when:

  • All cells of interest fit into a 20-quad
  • There is a relatively small number of living cells (because memory burden grows as O(living)
  • You are making a small number of steps at a time
  • Living cells are mostly still Lifes or period-two oscillators; the number of “active” Quad4s is relatively small

Gosper’s HashLife algorithm excels when:

  • Boards must be of unlimited size
  • Regularity in space — whether empty space or not — allows large regions to be deduplicated
  • You are making a large number of steps at a time
  • Regularity in time allows for big wins by caching and re-using steps we’ve seen already.
  • You’ve got a lot of memory! Because the caches are going to get big no matter what you do.

That’s why Gosper’s algorithm is so slow if you run in on the first few thousand generations of acorn; that evolution is very chaotic and so there are a lot of novel computations to do and comparatively less re-use. Once we’re past the chaotic period, things become very regular in both time and space, and we transition to a constant-time performance.


That is the last algorithm I’m going to present but I have one more thing to discuss in this series.

Next time on FAIC: we will finally answer the question I have been teasing all this time: are there patterns that grow quadratically? And how might our two best algorithms handle such scenarios?

 

Life, part 34

All right, we have our quad data structure, we know how to get and set individual elements, and we know how to display it. We’ve deduplicated it using memoization. How do we step it forward one tick? (Code for this episode is here.)

Remember a few episodes ago when we were discussing QuickLife and noted that if you have a 2-quad in hand, like these green ones, you can get the state of the blue 1-quad one tick ahead? And in fact we effectively memoized that solution by simply precomputing all 65536 cases.

The QuickLife algorithm memoized only the 2-quad-to-center-1-quad step algorithm; we’re going to do the same thing but with even more memoization. We have a recursively defined quad data structure, so it makes sense that the step algorithm will be recursive. We will use 2-quad-to-1-quad as our base case.

For the last time in this series, let’s write the Life rule:

private static Quad Rule(Quad q, int count)
{
  if (count == 2) return q;
  if (count == 3) return Alive;
  return Dead;
}

We’ll get all sixteen cells in the 2-quad as numbers:

private static Quad StepBaseCase(Quad q)
{
  Debug.Assert(q.Level == 2);
  int b00 = (q.NW.NW == Dead) ? 0 : 1;
  ... 15 more omitted ...

and count the neighbours of the center 1-quad:

  int n11 = b00 + b01 + b02 + b10 + b12 + b20 + b21 + b22;
  int n12 = b01 + b02 + b03 + b11 + b13 + b21 + b22 + b23;
  int n21 = b11 + b12 + b13 + b21 + b23 + b31 + b32 + b33;
  int n22 = b10 + b11 + b12 + b20 + b22 + b30 + b31 + b32;
  return Make(
    Rule(q.NW.SE, n11),
    Rule(q.NE.SW, n12),
    Rule(q.SE.NW, n21),
    Rule(q.SW.NE, n22));
}

We’ve seen this half a dozen times before. The interesting bit comes in the recursive step. The key insight is: for any n>=2, if you have an n-quad in hand, you can compute the (n-1) quad in the center, one tick ahead.

How? We’re going to use almost the same technique that we used in QuickLife. Remember in QuickLife the key was observing that if we had nine Quad2s in the old generation, we could compute a Quad3 in the new generation with sixteen steps on component Quad2s. The trick here is almost the same. Let’s draw some diagrams.

Suppose we have the 3-quad from the image above. We compute the next generation of its four component 2-quads; the green quads are current, the blue are stepped one ahead.

We can use a similar trick as we used with QuickLife to get the north, south, east, west and center 2-quads of this 3-quad, and move each of them ahead one step to get five more 1-quads. I’ll draw the original 3-quad in light green, and we can extract component 2-quads from it that I’ll draw in dark green. We then move each of those one step ahead to get the blue 1-quads.

That gives us this information:

We then make four 2-quads from those nine: and extract the center 1-quad from each using the Center function (source code below). I’ll just show the northwest corner; you’ll see how this goes. We make the light blue 2-quad out of four of the blue 1-quads, and then the center 1-quad of that thing is:

We do that four times and from those 1-quads we construct the center 2-quad moved one step ahead:

Summing up the story so far:

  • We can take a 2-quad forward one tick to make a 1-quad with our base case.
  • We’ve just seen here that we can use that fact to take a 3-quad forward one tick to make a 2-quad stepped forward one tick.
  • But nothing we did in the previous set of steps depended on having a 3-quad specifically. Assume that for some n >= 2 we can move an n-quad forward one tick to make an (n-1) quad; we have above an algorithm where we use that assumption and can move an (n+1)-quad forward to get an n-quad.

That is, we can move a 2-quad forward with our base case; moving a 3-quad forward requires the ability to move a 2-quad forward. Moving a 4-quad forward requires the ability to move a 3-quad forward, and so on.

As I’ve said many times on this blog, every recursive algorithm is basically the same. If we’re in the base case, solve the problem directly. If we’re not in the base case, break up the problem into finitely many smaller problems, solve each, and use the solutions to solve the larger problem.

Let’s write the code to move any n-quad for n >= 2 forward one tick.

We’ll need some helper methods that extract the five needed sub-quads, but those are easily added to Quad. (Of course these helpers are only valid when called on a 2-quad or larger.)

public Quad Center => Make(NW.SE, NE.SW, SE.NW, SW.NE);
public Quad N => Make(NW.NE, NE.NW, NE.SW, NW.SE);
public Quad E => Make(NE.SW, NE.SE, SE.NE, SE.NW);
public Quad S => Make(SW.NE, SE.NW, SE.SW, SW.SE);
public Quad W => Make(NW.SW, NW.SE, SW.NE, SW.NW);

And then I’ll make a static method that takes a Quad and returns the center stepped forward one tick. (Why not an instance method on Quad? We will see in a moment.)

private static Quad Step(Quad q)
{
  Debug.Assert(q.Level >= 2);
  Quad r;
  if (q.IsEmpty)
    r = Empty(q.Level - 1);
  else if (q.Level == 2)
    r = StepBaseCase(q);
  else
  {
    Quad q9nw = Step(q.NW);
    Quad q9n = Step(q.N);
    Quad q9ne = Step(q.NE);
    Quad q9w = Step(q.W);
    Quad q9c = Step(q.Center);
    Quad q9e = Step(q.E);
    Quad q9sw = Step(q.SW);
    Quad q9s = Step(q.S);
    Quad q9se = Step(q.SE);
    Quad q4nw = Make(q9nw, q9n, q9c, q9w);
    Quad q4ne = Make(q9n, q9ne, q9e, q9c);
    Quad q4se = Make(q9c, q9e, q9se, q9s);
    Quad q4sw = Make(q9w, q9c, q9s, q9sw);
    Quad rnw = q4nw.Center;
    Quad rne = q4ne.Center;
    Quad rse = q4se.Center;
    Quad rsw = q4sw.Center;
    r = Make(rnw, rne, rse, rsw);
  }
  Debug.Assert(q.Level == r.Level + 1);
  return r;
}

Well that was easy! We just do nine recursions and then reorganize the resulting nine one-tick-forward quads until we get the information we want, and return it. (I added a little easy out for the empty case, though strictly speaking that is not necessary.)

There are probably three things on your mind right now.

  • If we get a full quad-size smaller every time we step, we’re going to get down to a very small board very quickly!
  • QuickLife memoized the step-the-center-of-a-2-quad operation. Why aren’t we doing the same thing here?
  • Nine recursions is a lot; isn’t this going to blow up performance? Suppose we have an 8-quad; we do nine recursions on 7-quads, but each of those does nine recursions on 6-quads, and so on down to 3-quads. It looks like we are doing 9n-2 calls to the base case when stepping an n-quad forward one tick.

First things first.

When do we not care if we’re shrinking an n-quad down to an (n-1)-quad on step? When all living cells in the n-quad are already in the center (n-1)-quad. But that condition is easy to achieve! Let’s actually write our public step method, not just the helper that steps a quad. And heck, let’s make sure that we have more than enough empty space. Remember, empty space is super cheap. 

sealed class Gosper : ILife, IDrawScale, IReport
{
  private Quad cells;
  private long generation;
  ...
  public void Step()
  {
    Quad current = cells;

Make cells bigger until there are two “levels” of empty cells surrounding the center. (We showed Embiggen last time.) That way we are definitely not throwing away any living cells when we get a next state that is one size smaller:

    if (!current.HasAllEmptyEdges)
      current = current.Embiggen().Embiggen();
    else if (!current.Center.HasAllEmptyEdges)
      current = current.Embiggen();
    Quad next = Step(current);

We’ve stepped, so next is one size smaller than current. Might as well make it bigger too; that’s one fewer thing to deal with next time. Again, empty space is cheap.

  cells = next.Embiggen();
  generation += 1;
}

HasAllEmptyEdges is an easy helper method of Quad:

public bool HasAllEmptyEdges => 
  NW.NW.IsEmpty &&
  NW.NE.IsEmpty &&
  NE.NW.IsEmpty &&
  NE.NE.IsEmpty &&
  NE.SE.IsEmpty &&
  SE.NE.IsEmpty &&
  SE.SE.IsEmpty &&
  SE.SW.IsEmpty &&
  SW.SE.IsEmpty &&
  SW.SW.IsEmpty &&
  SW.NW.IsEmpty &&
  NW.SW.IsEmpty;

That was an easy problem to knock down. Second problem: QuickLife memoized the 2-quad-to-1-quad step algorithm and got a big win; shouldn’t we do the same thing?

Sure, we have a memoizer, we can do so easily. But… what about our third problem? We have a recursive step that is creating exponentially more work as the quad gets larger as it traverses our deduplicated tree structure.

Hmm.

It is recursing on a deduplicated structure, which means it is probably going to encounter the same problems several times. If we move a 3-quad forward one step to get a 2-quad, we’re going to get the same answer the second time we do the same operation on the same 3-quad. If we move a 4-quad forward one step to get a 3-quad, we will get the same answer the second time we do it. And so on. Let’s just memoize everything.

We’ll rename Step to UnmemoizedStep, create a third memoizer, and replace Step with:

private static Quad Step(Quad q) => 
  CacheManager.StepMemoizer.MemoizedFunc(q);

And now we have solved our second and third problems with one stroke.

Let’s run it! We’ll do our standard performance test of 5000 generations of acorn:

Algorithm           time(ms) size  Mcells/s 
Naïve (Optimized):   4000     8      82     
Abrash (Original)     550     8     596     
Stafford              180     8    1820     
QuickLife              65    20      ?      
Gosper v1            3700    60      ?

Oof.

It’s slow! Not as slow as the original naïve implementation, but just about.

Hmm.

That’s the time performance; what’s the memory performance? There’s a saying I’ve used many times; I first heard it from Raymond Chen but I don’t know if he coined it or was quoting someone else. “A cache without an expiration policy is called a memory leak”. Memory leaks can cause speed problems as well as memory problems because they increase burden on the garbage collector, which can slow down the whole system. Also, dictionaries are in theory O(1) access — that is, access time is the same no matter how big the cache gets — but theory and practice are often different as the dictionaries get very large.

How much memory are we using in this thing? The “empty” memoizer never has more than 61 entries, so we can ignore it. I did some instrumentation of the “make” and “step” caches; after 5000 generations of acorn:

  • the step and make caches both have millions of entries
  • half the entries were never read at all, only written
  • 97% of the entries were read fewer than twenty times
  • the top twenty most-read entries account for 40% of the total reads

This validates our initial assumption that there is a huge amount of regularity; the “unusual” situations recur a couple dozen times tops, and we spend most of our time looking at the same configurations over and over again.

This all suggests that we could benefit from an expiration policy. There are two things to consider: what to throw away, and when to throw it away. First things first:

  • An LRU cache seems plausible; that is, when you think it is time to take stuff out of the cache, take out some fraction of the stuff that has been Least Recently Used. However LRU caches involve making extra data structures to keep track of when something has been used; we do extra work on each step, and it seems like that might have a performance impact given how often these caches are hit.
  • The easiest policy is: throw it all away! Those 20 entries that make up 40% of the hits will be very quickly added back to the cache.

Let’s try the latter because it’s simple. Now, we cannot just throw it all away because we must maintain the invariant that Make agrees with Empty; that is, when we call Make with four empty n-quads and when we call Empty(n+1) we must get the same object out. But if we throw away the “make” cache and then re-seed it with the contents of the “empty” cache — which is only 61 entries, that’s easy — then we maintain that invariant.

When to throw it away? What we definitely do not want to happen is end up in a situation where we are throwing away stuff too often. We can make a very simple dynamically-tuned cache with this policy:

  • Set an initial cache size bound. 100K entries, 1M entries, whatever.
  • Every thousand generations, check to see if we’ve exceeded the cache size bound. If not, we’re done.
  • We’ve exceeded the bound. Throw away the caches, do a single step, and re-examine the cache size; this tells us the cache burden of doing one tick.
  • The new cache size bound is the larger of the current bound and twice the one-tick burden. That way if necessary the size bound gradually gets larger so we do less frequent cache resetting.

The code is straightforward; at the start of Step:

bool resetMaxCache = false;
if ((generation & 0x3ff) == 0)
{
  int cacheSize = CacheManager.MakeQuadMemoizer.Count + 
    CacheManager.StepMemoizer.Count;
  if (cacheSize > maxCache)
  {
    resetMaxCache = true;
    ResetCaches();
  }
}

“ResetCaches” throws away the step cache and resets the make cache to agree with the empty cache; I won’t bother to show it. At the end of Step:

if (resetMaxCache)
{
  int cacheSize = CacheManager.MakeQuadMemoizer.Count + 
    CacheManager.StepMemoizer.Count;
  maxCache = Max(maxCache, cacheSize * 2);
}

All right, let’s run it again!

Algorithm           time(ms) size  Mcells/s 
Naïve (Optimized):   4000     8      82     
Abrash (Original)     550     8     596     
Stafford              180     8    1820     
QuickLife              65    20      ?      
Gosper v2            4100    60      ?

It’s worse. Heck, it is worse than the naive algorithm!

Sure, the top twenty cache entries account for 40% of the hits, and sure, 97% of the entries are hit fewer than twenty times. But the statistic that is relevant here that I omitted is: the top many hundreds of cache entries account for 50% of the hits. We don’t have to rebuild just the top twenty items in the cache to start getting a win from caching again. We take a small but relevant penalty every time we rebuild the caches.

Sigh.

We could keep on trying to improve the marginal performance by improving our mechanisms. We could try an LRU cache, or optimize the caches for reading those top twenty entries, or whatever. We might eke out some wins. But maybe instead we should take a step back and ask if there’s an algorithmic optimization that we missed.


Next time on FAIC: There is an algorithmic optimization that we missed. Can you spot it?

Installing windows

Episode 34 will be delayed again — sorry! — because once again the time I had set aside for writing this weekend got consumed by a real-world task that could not wait. (I will try for Thursday of this week.)

Some friends who are moving had a handyman failure; as is often the case when renovating a house to be sold, they have a set of build dependencies that required this window to be replaced in a hurry in order to not slip the schedule for other renovations, so I volunteered to take care of it.

Yuck.

Living in a 112 year old house myself, I am used to doing archaeological investigations of the strange decisions made by previous owners. This window, though obviously an old single-paned window, did not look like it was original to the 120-year-old house. It was the wrong size for the rough opening; the hinges looked more modern than turn-of-the-century, and so on.

Sure enough, when disassembled there was a gap behind the trim that was insulated with crumpled newspapers from 1957. Oddly enough they were Pittsburgh newspapers from different days; perhaps the owners of the house in 1957 moved from Pittsburgh, replaced a window, and insulated the gaps with the packing paper they moved with? It’s a mystery.

Having zero haircuts since quarantine began has done wonders for my hair.

New window in and trimmed — obviously the paint will need to be redone but that’s why the window had to go in before the painters arrived.

And the interior needs a little more drywalling and priming before it is ready for painting, but it is 1000000x better than before at least.

The neighbours in the blue house apparently asked my friends for my contact information as they also have a window that needs replacing. I am quite chuffed. I had my friends pass along that I only do windows as a favour, but I would be happy to design them a programming language for hire should they need one of those.

Next time: Gosper’s algorithm, finally!

 

Implementing a full fence

Episode 34 of my ongoing series will be slightly delayed because I spent the time on the weekend I normally spend writing instead rebuilding one of my backyard fences.

I forgot to take a before picture, but believe me, it was ruinous when I bought the place in 1997 and to the point in 2020 where it was actually falling to pieces in a stiff breeze.

I replaced it with an identical design and materials:

…and divided and re-homed some sixteen dozen irises in the process.

It’s nice implementing something that requires no typing every now and then.

 

Life, part 33

Last time in this series we learned about the fundamental (and only!) data structure in Gosper’s algorithm: a complete quadtree, where every leaf is either alive or dead and every sub-tree is deduplicated by memoizing the static factory.

Suppose we want to make this 2-quad, from episode 32:

This is easy enough to construct recursively:

Quad q = Make(
  Make(Dead, Alive, Alive, Dead), // NW
  Empty(1), // NE
  Make(Dead, Dead, Alive, Alive), // SE
  Make(Dead, Alive, Alive, Dead)) // SW

Since Make is memoized, the NW and SW corners will be reference-equal.

But suppose we had this quad in hand and we wanted to change one of the cells from dead to alive — how would we do that? Since the data structure is immutable, a change produces a new object; since it is persistent, we re-use as much of the old state as possible. But because we re-use quads arbitrarily often, a quad cannot know its own location. 

The solution is: every time we need to refer to a specific location within a quad, we must also say what the coordinates are in the larger world of the lower left corner cell of that quad.

Let’s start with some easy stuff. Suppose we have a quad (this), the coordinate of its lower-left corner, and a point of interest (p). We wish to know: is p inside this quad or not?

(Source code for this episode is here.)

private bool Contains(LifePoint lowerLeft, LifePoint p) => 
  lowerLeft.X <= p.X && p.X < lowerLeft.X + Width &&
  lowerLeft.Y <= p.Y && p.Y < lowerLeft.Y + Width;

Easy peasy. This enables us to implement a getter; again, this is a method of Quad:

// Returns the 0-quad at point p
private Quad Get(LifePoint lowerLeft, LifePoint p)
{

If the point is outside the quad, assume it is dead.

  if (!Contains(lowerLeft, p))
    return Dead;

The point is inside the quad. Did we find the 0-quad we’re after? Return it!

  if (Level == 0)
    return this;

We did not find it, but it is around here somewhere. It must be in one of the four children, so figure out which one and recurse. Remember, we’ll need to compute the lower left corner of the quad we’re recursing on.

  long w = Width / 2;
  if (p.X >= lowerLeft.X + w)
  {
    if (p.Y >= lowerLeft.Y + w)
      return NE.Get(new LifePoint(lowerLeft.X + w, lowerLeft.Y + w), p);
    else
      return SE.Get(new LifePoint(lowerLeft.X + w, lowerLeft.Y), p);
  } 
  else if (p.Y >= lowerLeft.Y + w)
    return NW.Get(new LifePoint(lowerLeft.X, lowerLeft.Y + w), p);
  else
    return SW.Get(lowerLeft, p);
  }
}

Once we’ve seen the getter, it’s easier to understand the setter. The setter returns a new Quad:

private Quad Set(LifePoint lowerLeft, LifePoint p, Quad q)
{
  Debug.Assert(q.Level == 0);

If the point is not inside the quad, the result is no change.

  if (!Contains(lowerLeft, p))
    return this;

The point is inside the quad. If we are changing the value of a 0-quad, the result is the 0-quad we have in hand.

if (Level == 0)
  return q;

We need to recurse; you might think that the setter logic is going to be a mess right now, but actually it is very straightforward. Since setting a point that is not in range is an immediate identity, we can just recurse four times!

long w = Width / 2;
return Make(
  NW.Set(new LifePoint(lowerLeft.X, lowerLeft.Y + w), p, q),
  NE.Set(new LifePoint(lowerLeft.X + w, lowerLeft.Y + w), p, q),
  SE.Set(new LifePoint(lowerLeft.X + w, lowerLeft.Y), p, q),
  SW.Set(lowerLeft, p, q)) ;
}

That’s the internal logic for getting and setting a 0-quad inside a larger quad. I’ll also add public get and set methods that are just wrappers around these; the details are not particularly interesting.


What about drawing the screen? We haven’t talked about it for a while, but remember the interface that we came up with way back in the early episodes: the UI calls the Life engine with the screen rectangle in Life grid coordinates, and a callback that is called on every live cell in that rectangle:

 void Draw(LifeRect rect, Action<LifePoint> setPixel);

The details of “zooming in” to draw the pixels as squares instead of individual pixels was entirely handled by the UI, not by the Life engine, which should make sense.

We can quickly determine whether a given quad is empty, and therefore we can optimize the drawing engine to skip over all empty quads without considering their contents further, and that’s great. But there is another possibility here: because we can determine quickly if a quad is not empty, we could implement “zooming out”, so that every on-screen pixel represented a 1-quad or a 2-quad or whatever; we can zoom out arbitrarily far.

Let’s implement it! I’m going to add a new method to Quad that takes four things:

  • What is the coordinate address of the lower left corner of this quad?
  • What rectangle, in Life coordinates, are we drawing?
  • The callback
  • What is the level of the smallest quad we’re going to draw? This is the zoom factor.
private void Draw(
  LifePoint lowerLeft, 
  LifeRect rect, 
  Action<LifePoint> setPixel, 
  int scale)
{
  Debug.Assert(scale >= 0);

If this quad is empty then that’s easy; there’s nothing to draw.

  if (IsEmpty)
    return;

The quad is not empty. If this quad does not overlap the rectangle, there’s nothing to draw. Unfortunately I chose to make the rectangle be “top left corner, width, height” but we are given the bottom left corner, so I have a small amount of math to do.

  if (!rect.Overlaps(new LifeRect(
      lowerLeft.X, lowerLeft.Y + Width - 1, Width, Width)))
    return;

(“Do these rectangles overlap?” is of course famously an interview problem; see the source code for my implementation.)

If we made it here we have a non-empty quad that overlaps the rectangle. Is this the lowest level we’re going to look at? Then set that pixel.

  if (Level <= scale)
  {
    setPixel(lowerLeft);
    return;
  }

We are not at the lowest level yet. Simply recurse on the four children; if any of them are empty or not overlapping, they’ll return immediately without doing more work.

  long w = Width / 2;
  NW.Draw(new LifePoint(lowerLeft.X, lowerLeft.Y + w), rect, setPixel, scale);
  NE.Draw(new LifePoint(lowerLeft.X + w, lowerLeft.Y + w), rect, setPixel, scale);
  SE.Draw(new LifePoint(lowerLeft.X + w, lowerLeft.Y), rect, setPixel, scale);
  SW.Draw(lowerLeft, rect, setPixel, scale);
}

That’s the internal logic; there is then a bunch of public wrapper methods around it that I will omit, and a new interface IDrawScale. We now have enough gear that we can start implementing our actual engine:

sealed class Gosper : ILife, IReport, IDrawScale
{
  private Quad cells;
  public Gosper()
  {
    Clear();
  }
  public void Clear()
  {
    cells = Empty(9);
  }
  public bool this[LifePoint p]
  {
    get => cells.Get(p) != Dead;

That’s all straightforward. But what about the setter? We’ve made a 9-quad, but what if we try to set a point outside of it? Fortunately it is very cheap to expand a 9-quad into a 10-quad, and then into an 11-quad, and so on, as we’ve seen.

    set
    {
      while (!cells.Contains(p))
        cells = cells.Embiggen();
      cells = cells.Set(p, value ? Alive : Dead);
    }
  }

The implementation of Embiggen is just what you would expect:

public Quad Embiggen()
{
  Debug.Assert(Level >= 1);
  if (Level >= MaxLevel)
    return this;
  Quad q = Empty(this.Level - 1);
  return Make(
    Make(q, q, NW, q), Make(q, q, q, NE),
    Make(SE, q, q, q), Make(q, SW, q, q));
}

And the rest of the methods of our Gosper class are uninteresting one-liners that just call the methods we’ve already implemented.


I’ve updated the UI to support “zoom out” functionality and created a Life engine that does not step, but just displays a Quad. Previously we could display a zoomed-in board:

or one pixel per cell:

But now we can display one pixel per 1-quad:

or one pixel per 2-quad, or whatever; we can zoom out arbitrarily far. If any cell in the quad is on, the pixel is on, which seems reasonable. This will let us much more easily visualize patterns that are larger than the screen.


Next time on FAIC: How do we step a quad forward one tick?

Approximate results may vary

Part 33 of my ongoing series is coming but I did not get all the code written that I wanted to this week, so it will be delayed. In the meanwhile:


Living in Canada as a child, of course I grew up learning the metric system (with some familiarity with the imperial and US systems of course). If you want to know how many milliliters of liquid a cubic box holds, you just compute the volume of the box in cubic centimeters and you’re done, because a milliliter is by definition the same volume as a cubic centimeter.

Despite having lived in Seattle for over 25 years, I still sometimes do not have an intuitive sense of conversions of US units; for a particular project I knew that I needed an amount of liquid equal to about the volume of a two-inch cube, but the bottle was measured in fluid ounces. Fortunately we have this operation in every browser:

Thanks Bing for those important eight digits after the decimal place there, manufactured from the zero digits after the decimal place of the input. For context, that extra 0043 on the end represents a volume equivalent to roughly the size of a dozen specs of microscopic dust.

But the punchline that made me LOL was “for an approximate result, divide the volume by 1.8046875”, because of course when I am quickly approximating the volume of eight cubic inches in fluid ounces, the natural operation that immediately comes to mind is divide by 1.8046875.

I have some questions.

That was Bing; what does Google do with the same query?

OK, that’s an improvement in that the amount of precision is still unnecessary, but not outright absurd. But the other problems are all the same.

Why division? Maybe it is just me, but I find division by uneven quantities significantly harder mental arithmetic than multiplication; assuming we want the over-precision, would it not be better to say “for an approximate result, multiply the volume by 0.5541125″? 

Why say “to approximate…” and then give an absurd amount of precision in the conversion factor? Approximation is by definition about deliberately not computing an over-precise value.

Surely the right way to say this is “for an approximate result, multiply the volume by 5/9” or even better, “divide by two“. When I saw “divide by 1.8046875” the first thing I thought after “wow that’s so over-precise” was “1.8 is 18/10 is 9/5, so multiply by 5/9“.

I’m going to get there eventually; software can shorten that journey. And I’m going to remember that 5/9ths of a cubic inch is a fluid ounce much more easily than I remember to divide by 1.8046875.

Once you start to see this pattern of over-precision in conversions, it’s like the FedEx arrow: you can’t stop seeing it. Let’s ask the browser how much does an American robin weigh?


64.8 grams. An underweight robin is not 64.5 grams, and not 65.0 grams but 64.8 grams.

To be fair, it looks like this over-precision was the fault of a human author (and their editor) not thinking clearly about how to communicate the fact at hand, rather than bad software this time; if you’re converting “two and a third ounces” to grams it would be perfectly reasonable to round to 65, or even 60. (That third of an ounce is already suspect; surely “two to three ounces” is just fine.) Most odd though is that the computations are not even correct! 2 and 1/3rd ounces is 66.15 grams, and 3 ounces is 85.05 grams, making it rather mysterious where the extra few hundred milligrams went.

I was wondering how many earthworms a robin would have to eat to make up a discrepancy of 0.2 grams. A largish earthworm has got to weigh on the order of a gram, right?

Wow! (For my metric readers out there: 1.5 pounds is 680.388555 grams according to Bing.)

Again: what the heck, Bing? I did not ask for “world’s largest earthworm” or “unusually large earthworms” or even “Australian earthworms”. You know where I live, Bing. (And Google search does no better.)

For some reason I am reminded of Janelle Shane’s “AI Weirdness” tweets; the ones about animal facts are particularly entertaining. These earthworm facts at least have the benefit of being both interesting and correct, but this is hardly the useful result about normal garden-variety annelids that I wanted.

I am also reminded of my favourite animal fact: the hippopotamus can jump higher than a house. It sounds impressive until you remember that houses can’t jump at all.

Obviously all these issues are silly and unimportant, which is why I chose them for my fun-for-Friday blog. And the fact that I can type “8 cubic inches in oz” into my browser or say “how much does a robin weigh?” into a smart speaker and instantly get the answer is already a user interface triumph; I don’t want to minimize that great work. But there is still work to do! Unwarranted extra precision is certainly not the worst sort of fallacious reasoning we see on the internet, but it is one of the most easily mitigated by human-focused software design. I’d love to see improvements to these search functions that show even more attention to what the human user really needs.

Life, part 32

All right, after that excessively long introduction let’s get into Gosper’s algorithm, also known as “HashLife” for reasons that will become clear quite soon. Since the early days of this series I’ve mostly glossed over the code that does stuff like changing individual cells in order to load in a pattern, or the code that draws the UI, but Gosper’s algorithm is sufficiently different that I’m going to dig into every part of the implementation.

The first key point to understand about Gosper’s algorithm is that it only has one data structure. One way to describe it is:

A quad is an immutable complete tree where every non-leaf node has exactly four children which we call NE, NW, SE, SW, and every leaf is either alive or dead.

However the way I prefer to describe it is via a recursive definition:

  • A 0-quad is either alive or dead.
  • An n-quad for n > 0 is four (n-1) quads.

Let’s write the code!

sealed class Quad
{
  public const int MaxLevel = 60;

Because there will be math to do on the width of the quad and I want to keep doing the math in longs instead of BigIntegers, I’m going to limit us to a 60-quad, max. Just because it is a nice round number less than the 64 bits we have in a long.

Now, I know what you’re thinking. “Eric, if we built a monitor with a reasonable pixel density to display an entire 60-quad it would not fit inside the orbit of Pluto and it would have more mass than the sun.” A 60-quad is big enough for most purposes. I feel good about this choice. However I want to be clear that in principle nothing is stopping us from doing math in a larger type and making arbitrarily large quads.

Here are our two 0-quads:

public static readonly Quad Dead = new Quad();
public static readonly Quad Alive = new Quad();

And for the non-0-quads, the child (n-1) quads:

public Quad NW { get; }
public Quad NE { get; }
public Quad SE { get; }
public Quad SW { get; }

We will need to access the level and width of a quad a lot, so I’m going to include the level in every quad. The width we can calculate on demand.

public int Level { get; }
public long Width => 1L << Level;

We’ll need some constructors. There’s never a need to construct 0-quads because we have both of them already available as statics. For reasons which will become apparent later, we have a public static factory for the rest.

public static Quad Make(Quad nw, Quad ne, Quad se, Quad sw)
{
  Debug.Assert(nw.Level == ne.Level);
  Debug.Assert(ne.Level == se.Level);
  Debug.Assert(se.Level == sw.Level);
  return new Quad(nw, ne, se, sw);
}
private Quad() => Level = 0;
private Quad(Quad nw, Quad ne, Quad se, Quad sw)
{
  NW = nw;
  NE = ne;
  SE = se;
  SW = sw;
  Level = nw.Level + 1;
}

We’re going to need to add a bunch more stuff here, but this is the basics.

Again I know what you’re probably thinking: this is bonkers. In QuickLife, a 3-quad was an eight byte value type. In my astonishingly wasteful implementation so far, a single 0-quad is a reference type, so it is already eight bytes for the reference, and then it contains four more eight byte references and a four byte integer that is never more than 60. How is this ever going to work?

Through the power of persistence, that’s how. As I’ve discussed many times before on this blog, a persistent data structure is an immutable data structure where, because every part of it is immutable, you can safely re-use portions of it as you see fit. You therefore save on space.

Let’s look at an example. How could we represent this 2-quad?

Remember, we only have two 0-quads and we cannot make any more. The naïve way would be to make this tree:

But because every one of these objects is immutable, we could instead make this tree, which has one fewer object allocation:

This is the second key insight to understanding how Gosper’s algorithm works: it uses a relatively enormous data structure for each cell, but it achieves compression through deduplication:

  • There are only two possible 0-quads, and we always re-use them.
  • There are 16 possible 1-quads. We could just make 16 objects and re-use them.
  • There are 65536 possible 2-quads, but the vast majority of them we never see in a given Life grid. The ones we do see, we often see over and over again. We could just make the ones we do see, and re-use them.

The same goes for 3-quads and 4-quads and so on. There are exponentially more possibilities, and we see a smaller and smaller fraction of them in any board. Let’s memoize the Make function!

We’re going to make heavy use of memoization in this algorithm and it will have performance implications later, so I’m going to make a relatively fully-featured memoizer whose behaviour can be analyzed:

sealed class Memoizer<A, R>
{
  private Dictionary<A, R> dict; 
  private Dictionary<A, int> hits;
  private readonly Func<A, R> f;
  [Conditional("MEMOIZER_STATS")]
  private void RecordHit(A a)
  {
    if (hits == null)
      hits = new Dictionary<A, int>();
    if (hits.TryGetValue(a, out int c))
      hits[a] = c + 1;
    else
      hits.Add(a, 1);
  }

  public R MemoizedFunc(A a)
  { 
    RecordHit(a); 
    if (!dict.TryGetValue(a, out R r)) 
    { 
      r = f(a); 
      dict.Add(a, r); 
    } 
    return r; 
  }
  public Memoizer(Func<A, R> f)
  {
    this.dict = new Dictionary<A, R>();
    this.f = f;
  }
  public void Clear(Dictionary<A, R> newDict = null)
  {
    dict = newDict ?? new Dictionary<A, R>();
    hits = null;
  }
  public int Count => dict.Count;
  public string Report() => 
    hits == null ? "" :
    string.Join("\n", from v in hits.Values 
                      group v by v into g 
                      select $"{g.Key},{g.Count()}");
}

The core logic of the memoizer is the same thing I’ve presented many times on this blog over the years: when you get a call, check to see if you’ve memoized the result; if you have not, call the original function and memoize the result, otherwise just return the result.

I’ve added a conditionally-compiled hit counter and a performance report that tells me how many items were hit how many times; that will give us some idea of the load that is being put on the memoizer, and we can then tune it.

Later in this series we’re going to need to “reset” a memoizer and optionally we’ll need to provided “pre-memoized” state, so I’ve added a “Clear” method that optionally takes a new dictionary to use.

The memoizer for the “Make” method can be static, global state so I’m going to make a helper class for the cache. (I did not need to; this could have been a static field of Quad. It was just convenient for me while I was developing the algorithm to put the memoizers in one central location.)

static class CacheManager
{
  public static Memoizer<(Quad, Quad, Quad, Quad), Quad> MakeQuadMemoizer { get; set; }
}

We’ll initialize it when we create our first Quad:

static Quad()
{
  CacheManager.MakeQuadMemoizer = 
    new Memoizer<(Quad, Quad, Quad, Quad), Quad>(UnmemoizedMake);
}

And we’ll redo the Make static factory:

private static Quad UnmemoizedMake((Quad nw, Quad ne, Quad se, Quad sw) args)
{
  Debug.Assert(args.nw.Level == args.ne.Level);
  Debug.Assert(args.ne.Level == args.se.Level);
  Debug.Assert(args.se.Level == args.sw.Level);
  return new Quad(args.nw, args.ne, args.se, args.sw);
}

public static Quad Make(Quad nw, Quad ne, Quad se, Quad sw) =>
  CacheManager.MakeQuadMemoizer.MemoizedFunc((nw, ne, se, sw));

All right, we have memoized construction of arbitrarily large quads. A nice consequence of this fact is that all quads can be compared for equality by reference equality. In particular, we are going to do two things a lot:

  • Create an arbitrarily large empty quad
  • Ask “is this arbitrarily large quad an empty quad”?

We’re on a memoization roll here, so let’s keep that going and add a couple more methods to Quad, and another memoizer to the cache manager (not shown; you get how it goes.)

private static Quad UnmemoizedEmpty(int level)
{
  Debug.Assert(level >= 0);
  if (level == 0)
    return Dead;
  var q = Empty(level - 1);
  return Make(q, q, q, q);
}

public static Quad Empty(int level) => 
  CacheManager.EmptyMemoizer.MemoizedFunc(level);

public bool IsEmpty => this == Empty(this.Level);

The unmemoized “construct an empty quad” function can be as inefficient as we like; it will only be called once per level. And now we can quickly tell if a quad is empty or not. Well, relatively quickly; we have to do a dictionary lookup and then a reference comparison.

What then is the size in memory of an empty 60-quad? It’s just 61 objects! The empty 1-quad refers to Dead four times, the empty 2-quad refers to the empty 1-quad four times, and so on.

Suppose we made a 3-quad with a single glider in the center; that’s a tiny handful of objects. If you then wanted to make a 53-quad completely filled with those, that only increases the number of objects by 50. Deduplication is super cheap with this data structure — provided that the duplicates are aligned on boundaries that are a power of two of course.

A major theme of this series is: find the characteristics of your problem that admit to optimization and take advantage of those in pursuit of asymptotic efficiency; don’t worry about the small stuff. Gosper’s algorithm is a clear embodiment of that principle. We’ve got a space-inefficient data structure, we’re doing possibly expensive dictionary lookups all over the show; but plainly we can compress down grids where portions frequently reoccur into a relatively small amount of memory at relatively low cost.


Next time on FAIC: There are lots more asymptotic wins to come, but before we get into those I want to explore some mundane concerns:

  • If portions of the data structure are reused arbitrarily often then no portion of it can have a specific location. How are we going to find anything by its coordinates?
  • If the data structure is immutable, how do we set a dead cell to alive, if, say, we’re loading in a pattern?
  • How does the screen drawing algorithm work? Can we take advantage of the simplicity of this data structure to enable better zooming in the UI?

 

 

Socially distant abbreviated summer vacation

Normally this time of year I would be visiting friends and family in Canada, but obviously that’s impossible right now. Instead we took a long weekend at a rental on Bainbridge Island and strolled around some parks in a socially distant manner instead.

The rental had a balcony at tree level, so at least I got to indulge in my vacation pastime of taking pictures of birds. There were a lot of common feeder birds in the neighbourhood — house finches, house sparrows, a few different species of chickadees, that sort of thing — and I got a few shots in. This little female Rufous hummingbirb posed for glamour shots in a birch tree all day long:

(Click on any picture for a larger image)

She was vexed that a female Anna’s hummingbird kept horning in on her territory, hence her ticked-off expression there.

This super fuzzy juvenile black-capped chickadee also spent days posing:

despite being harassed all day long by the noisy gang of teenager Stellar jays on the rooftop next door.

And of course once again I was completely incapable of taking a non-blurry picture of a belted kingfisher despite many attempts.

Sigh. Better luck next time.

Life, part 31

Today we will finish off our implementation of Hensel’s QuickLife algorithm, rewritten in C#. Code for this episode is here.

Last time we saw that adding change tracking is an enormous win, but we still have not got an O(changes) solution in time or an O(living) solution in space. What we really want to do is (1) avoid doing any work at all on stable Quad4s; only spend processor time on the active ones, and (2) deallocate all-dead Quad4s, and (3) grow the set of active Quad4s when activity reaches the current borders.

We need more state bits, but fortunately we only need a small number; so small that I’m not even going to bit twiddle them. (The original Java implementation did, but it also used some of those bits for concerns such as optimizing the display logic.) Recall that we are tracking 32 regions of the 8 Quad3s in a Quad4 to determine if they are active, stable or dead. It should come as no surprise that we’re going to do the same thing for a Quad4, and that once again we’re going to maintain state for both the even and odd cycles:

enum Quad4State
{
  Active,
  Stable,
  Dead
}

And then in Quad4:

public Quad4State State { get; set; } 
public Quad4State EvenState { get; set; }
public Quad4State OddState { get; set; }
public bool StayActiveNextStep { get; set; }

The idea here is:

  • All Quad4s are at all times in exactly one of three buckets: active, stable and dead. Which bucket is represented by the State property.
  • If a Quad4 is active then it is processed on each tick.
  • If a Quad4 is stable then we draw it but do not process it on each tick.
  • If a Quad4 is dead then we neither process nor draw it, and eventually we deallocate it. (We do not want to deallocate too aggressively in case it comes back to life in a few ticks.)

How then do we determine when an active Quad4 becomes stable or dead?

  • If the “stay active” flag is set, it stays active no matter what.
  • The odd and even cycles each get a vote on whether an active Quad4 should become stable or dead.
  • If both vote for dead, it becomes dead.
  • If one votes for stable and the other votes for stable or dead, it becomes stable. (Exercise: under what circumstances is a Quad4 dead on the odd cycles but stable on the even?)
  • If one or both vote for active, it stays active.

How do we determine if a stable (or dead but not yet deallocated) Quad4 becomes active? That’s straightforward: if an active Quad4 has an active edge that abuts a stable Quad4, the stable one becomes active. Or, if there is no Quad4 there at all, we allocate a new one and make it active; this is how we achieve our goal of a dynamically growing board.

You might be wondering why we included a “stay active” flag. There were lots of things about this algorithm that I found difficult to understand at first, but this took the longest for me to figure out, oddly enough.

This flag means “there was activity on the edge of a neighbouring Quad4 recently”. There are two things that could go wrong in that situation that we need to prevent. First, we could have an active Quad4 that is about to become stable or dead, but it has activity on a border that should cause it to remain active for processing. Second, we could have a stable (or dead) Quad4 that has just been marked as active because there is activity on a bordering Quad4, and we need to ensure that it stays active even though it wants to go back to being stable (or dead).


The easiest task to perform is to keep each Quad4 in one of three buckets. The original implementation did so by ensuring that every Quad4 was on exactly one of three double-linked lists, and I see no reason to change that. I have, however, encapsulated the linking and unlinking into a helper class:

interface IDoubleLink<T> where T : class
{
  T Prev { get; set; }
  T Next { get; set; }
}

sealed class DoubleLinkList<T> : IEnumerable<T> 
  where T : class, IDoubleLink<T>
{
  public int Count { get; private set; }
  public void Add(T item) { ... }
  public void Remove(T item) { ... }
  public void Clear() { ... }
}

I won’t go through the implementation; it is more or less the standard double-linked list implementation you know from studying for boring technical interviews. The only unusual thing about it is that I’ve ensured that you can remove the current item from a list safely even while the list is being enumerated, because we will need to do that.

All the Quad4-level state manipulation will be done by our main class; we’ll add some lists to it:

sealed class QuickLife : ILife, IReport
{
  private readonly DoubleLinkList<Quad4> active = new DoubleLinkList<Quad4>();
  private readonly DoubleLinkList<Quad4> stable = new DoubleLinkList<Quad4>();
  private readonly DoubleLinkList<Quad4> dead = new DoubleLinkList<Quad4>();
  private Dictionary<(short, short), Quad4> quad4s;
  private int generation;

I’ll skip the initialization code and whatnot.

We already have a method which allocates Quad4s and ensures their north/south, east/west, northwest/southeast references are initialized. We will need a few more helper functions to encapsulate some operations such as: make an existing Quad4 active, and if it does not exist, create it. Or, make an active Quad4 into a stable Quad4. They’re for the most part just updating lists and state flags:

private Quad4 EnsureActive(Quad4 q, int x, int y)
{
  if (q == null)
    return AllocateQuad4(x, y);
  MakeActive(q);
  return q;
}
private void MakeActive(Quad4 q)
{
  q.StayActiveNextStep = true;
  if (q.State == Active) 
    return;
  else if (q.State == Dead)
    dead.Remove(q);
  else
    stable.Remove(q);
  active.Add(q);
  q.State = Active;
}
private void MakeDead(Quad4 q)
{
  Debug.Assert(q.State == Active);
  active.Remove(q);
  dead.Add(q);
  q.State = Dead;
}
private void MakeStable(Quad4 q)
{
  Debug.Assert(q.State == Active);
  active.Remove(q);
  stable.Add(q);
  q.State = Stable;
}

Nothing surprising there, except that as I mentioned before, when you force an already-active Quad4 to be active, it sets the “stay active for at least one more tick” flag.

There is one more bit of list management mechanism we should consider before getting into the business logic: when and how do dead Quad4s get removed from the dead list and deallocated? The how is straightforward: run down the dead list, orphan all of the Quad4s by de-linking them from every living object, and the garbage collector will eventually get to them:

private void RemoveDead()
{
  foreach(Quad4 q in dead)
  {
    if (q.S != null) 
      q.S.N = null;
    ... similarly  for N, E, W, SE, NW
    quad4s.Remove((q.X, q.Y));
  }
  dead.Clear();
}

This orphans the entire dead list; the original implementation had a more sophisticated implementation where it would keep around the most recently dead on the assumption that they were the ones most likely to come back.

We have no reason to believe that this algorithm’s performance is going to be gated on spending a lot of time deleting dead nodes, so we’ll make an extremely simple policy; every 128 ticks we’ll check to see if there are more than 100 Quad4s on the dead list.

private bool ShouldRemoveDead => 
  (generation & 0x7f) == 0 && dead.Count > 100;
public void Step()
{
  if (ShouldRemoveDead) 
    RemoveDead();
  if (IsOdd)
    StepOdd();
  else
    StepEven();
  generation += 1;
}

All right, that’s the mechanism stuff. By occasionally pruning away all-dead Quad4s we attain O(living) space usage. But how do we actually make this thing both fast and able to dynamically add new Quad4s on demand?

In keeping with the pattern of practice so far, we’ll write all the code twice, once for the even cycle and once, slightly different, for the odd cycle. I’ll only show the even cycle here.

Remember our original O(cells) prototype stepping algorithm:

private void StepEven()
{
  foreach (Quad4 q in quad4s)
    q.StepEven();
}

We want to (1) make this into an O(changes) algorithm, (2) detect stable or dead Quad4s currently on the active list, and (3) expand the board by adding new Quad4s when the active region reaches the current edge. We also have (4) a small piece of bookkeeping from earlier to deal with:

private void StepEven()
{
  foreach (Quad4 q in active)      // (1) O(changes)
  {
    if (!RemoveStableEvenQuad4(q)) // (2) remove and skip stable/dead
    {
      q.StepEven();
      MakeOddNeighborsActive(q);   // (3) expand
    }
    q.StayActiveNextStep = false;  // (4) clear flag
  }
}

The first one is straightforward; we now only loop over the not-stable, not-dead Quad4s, so this is O(changes), and moreover, remember that we consider a Quad4 to be stable if both its even and odd generations are stable, so we are effectively skipping all computations of Quad4s that contain only still Lifes and period-two oscillators, which is a huge win.

The fourth one is also straightforward: if the “stay active for at least one step” flag was on, well, we made it through one step, so it can go off. If it was already off, it still is.

The interesting work comes in removing stable Quad4s, and expanding the board on the active edge. To do the former, we will need some more helpers that answer questions about the Quad4 and its neighbors; this is a method of Quad4:

public bool EvenQuad4OrNeighborsActive =>
  EvenQuad4Active ||
  (S != null && S.EvenNorthEdgeActive) ||
  (E != null && E.EvenWestEdgeActive) ||
  (SE != null && SE.EvenNorthwestCornerActive);

This is the Quad4 analogue of our earlier method that tells you if any of a 10×10 region is active; this one tells you if an 18×18 region is active, and for the same reason: because that region entirely surrounds the new shifted-to-the-southeast Quad4 we’re computing. We also have the analogous methods to determine if that region is all stable or all dead; I’ll omit showing them.

Let’s look at our “remove a stable/dead Quad4 from the active list” method. There is some slightly subtle stuff going on here. First, if the Quad4 definitely can be skipped for this generation because it is stable or dead, we return true. However, that does not guarantee that the Quad4 was removed from the active list! Second, remember that our strategy is to have both the even and odd cycles “vote” on what should happen:

private bool RemoveStableEvenQuad4(Quad4 q)
{
  if (q.EvenQuad4OrNeighborsActive)
  {
    q.EvenState = Active;
    q.OddState = Active;
    return false;
  }

If anything in the 18×18 region containing this Quad4 or its relevant neighbouring Quad4s are active, we need to stay active. We set the votes of both even and odd cycle back to active if they were different.

If nothing is active then it must be stable or dead. Is this 18×18 region dead? (Remember, we only mark it as dead if it is both dead and stable.)

  if (q.EvenQuad4AndNeighborsAreDead)
  { 
    q.EvenState = Dead;
    q.SetOddQuad4AllRegionsDead();
    if (!q.StayActiveNextStep && q.OddState == Dead)
      MakeDead(q);
  }

Let’s go through each line here in the consequence of the if:

  • Everything in an 18×18 region is dead and stable. The even cycle votes for death.
  • We know that an 18×18 region is all dead and stable; that region entirely surrounds the odd Quad4. We therefore know that the odd Quad4 will be all dead on the next tick if it is not already, so we get aggressive and set all its “region dead” bits now.
  • If we’re forcing this Quad4 to stay active then it stays active; however, even is still voting for death! We’ll get another chance to kill this Quad4 on the next tick.
  • By that same logic, if the odd cycle voted for death on the previous tick but the Quad4 stayed active for some reason then the odd cycle is still voting for death now. If that’s true then both cycles are voting for death and the Quad4 gets put on the dead list.

We then have nearly identical logic for stability; the only difference is that if one cycle votes for stability, it suffices for the other cycle to vote for stability or death:

  else
  {
    q.EvenState = Stable;
    q.SetOddQuad4AllRegionsStable();
    if (!q.StayActiveNextStep && q.OddState != Active)
      MakeStable(q);
  }
  return true;
}

And finally: how do we expand into new space? That is super easy, barely an inconvenience. If we have just processed an active Quad4 on the even cycle then we’ve created a new odd Quad4 and in doing so we’ve set the activity bits for the 16 regions in the odd Quad4. If the south 16×2 edge of the odd Quad4 is active then the Quad4 to the south must be activated, and so on:

private void MakeOddNeighborsActive(Quad4 q)
{
  if (q.OddSouthEdgeActive)
    EnsureActive(q.S, q.X, q.Y - 1);
  if (q.OddEastEdgeActive)
    EnsureActive(q.E, q.X + 1, q.Y);
  if (c.OddSoutheastCornerActive)
    EnsureActive(q.SE, q.X + 1, q.Y - 1); 
}

Once again we are taking advantage of the fact that the even and odd generations are offset by one cell; we only have to expand the board on two sides of each Quad4 during each tick, instead of all four. When we’re completing an even cycle we check for expansion to the south and east for the upcoming odd cycle; when we’re completing an odd cycle we check for expansion to the north and west for the upcoming even cycle.

It’s a little hard to wrap your head around, but it all works. This “staggered” property looked like it was going to be a pain when we first encountered it, but it is surprisingly useful; that insight is one of the really smart things about this algorithm.

There is a small amount of additional state management code I’ve got to put here and there but we’ve hit all the high points; see the source code for the exact changes if you are curious.


And that is that! Let’s take it for a spin and run our usual 5000 generations of “acorn”.

Algorithm           time(ms) size  Mcells/s bits/cell O-time
Naïve (Optimized):   4000     8      82     8/cell     cells
Abrash (Original)     550     8     596     8/cell     cells
Stafford              180     8    1820     5/cell     change
Sparse array         4000    64      ?   >128/living   change
Proto-QuickLife 1     770     8     426     4/cell     cells
Proto-QuickLife 2     160     8    2050     4/cell     cells
QuickLife              65    20      ?      5/living*  change

WOW!

We have an O(changes) in time and O(living) in space algorithm that maintains a sparse array of Quad4s that gives us a 20-quad to play with, AND it is almost three times faster than Stafford’s algorithm on our benchmark!

Our memory usage is still pretty efficient; we are spending zero memory on “all dead” Quad4s with all dead neighbours. We’ve added more state to each Quad4, so now for active and stable Quad4s we’re spending around five bits per cell; same as Stafford’s algorithm. (Though as I noted in my follow-up, he did find ways to get “maintain a count of living neighbours” algorithms down to far fewer bits per cell.) I added an asterisk to the table above because of course an active or stable Quad4 will contain only 50% or fewer living cells, so this isn’t quite right, but you get the idea.

Again, it is difficult to know how to characterize “cells per second” for sparse array approaches where we have an enormous board that is mostly empty space that costs zero to process, so I’ve omitted that metric.


If you ran the code on your own machine you probably noticed that I added counters to the user interface to give live updates of the size of the active, stable and dead lists. Here’s a graph of the first 6700 generations of acorn (click on the image for a larger version.)

You can really see how the pattern evolves from this peek into the workings of the algorithm!

  • We start with a small number of active Quad4s; soon small regions of stability start to appear as the pattern spreads out and leaves still Lifes and period-two oscillators in its wake.
  • The number of dead Quad4s remains very small right until the first glider escapes; from that moment on we have one or more gliders shooting off to infinity. In previous implementations they hit the wall of death, but now they are creating new active Quad4s in their bow shocks, and leaving dead Quad4s in their wakes.
  • The stable count steadily grows as the active region is a smaller and smaller percentage of the total. Around the 5000th generation everything except the escaped gliders is stable, and we end up with around 100 stable Quad4s and 20 active Quad4s for the gliders.
  • The action of our trivial little “garbage collector” is apparent here; we throw away the trash only when there are at least 100 dead Quad4s in the list and we are on a generation divisible by 128, so it is unsurprising that we have a sawtooth that throws away a little over 100 dead Quad4s every 128 cycles.
  • The blue line is proportional to time taken per cycle, because we only process active Quad4s.
  • The sum of all three lines is proportional to total memory used.

That finishes off our deep dive into Alan Hensel’s QuickLife algorithm. I was quite intimidated by this algorithm when I first read the source code, but once you name every operation and reorganize the code into functional areas it all becomes quite clear. I’m glad I dug into it and I learned a lot.


Coming up on FAIC:

We’ve seen a lot of varied ways to solve the problem of simulating Life, and there are a pile more in my file folder of old articles from computer magazines that we’re not going to get to in this series. Having done this exploration into many of them and skimmed a lot more, two things come to mind.

First, so far they all feel kind of the same to me. There is some sort of regular array of data, and though I might layer object-oriented features on top of it to make the logic easier to follow or better encapsulated, fundamentally we’re doing procedural work here. We can be more or less smart about what work we can avoid or precompute, and thereby eliminate or move around the costs, but the ideas are more or less the same.

Second, every optimization we’ve done increases the amount of redundancy, mutability, bit-twiddliness, and general difficulty of understanding the algorithm.

Gosper’s algorithm stands in marked contrast.

  • There is no “array of cells” at all
  • The attack on the problem is purely functional programming; there is very little state mutation.
  • There is no redundancy. In the Abrash, Stafford and Hensel algorithms we had ever-increasing amounts of redundant state that had to be carefully kept in sync with the board state. In Gosper’s algorithm, there is board state and nothing else.
  • No attempt whatsoever is made to make individual cells smaller in memory, but it can represent grids a trillion cells on a side with a reasonable amount of memory.
  • It can compute trillions of ticks per second on quadrillion-cell grids on machines that only do billions of operations per second.
  • Though there are plenty of tricky details to consider in the actual implementation, the core algorithm is utterly simple and elegant. The gear that does the optimization of the algorithm uses off-the-shelf parts and completely standard, familiar functional programming techniques. There is no mucking around with fiddly region change tracking, or change lists, or bit sets, or any of those mechanisms.
  • And extra bonus, the algorithm makes “zoom in or out arbitrarily far” in the UI super easy, which is nice for large boards.

This all sounds impossible. It is not an exaggeration to say that learning about this algorithm changed the way I think about the power of functional programming and data abstraction, and I’ve been meaning to write a blog about it for literally over a decade.

It will take us several episodes to get through it:

  • Next time on FAIC we’ll look at the core data structure. We’ll see how we can compress large boards down to a small amount of space.
  • Then we’ll closely examine the “update the UI” algorithm and see how we can get a nice new feature.
  • After that we’ll build the “step forward one generation algorithm” and do some experiments with it.
  • Finally, we’ll discover the key insight that makes Gosper’s algorithm work: you can enormously compress time if you make an extremely modest addition of extra space.
  • We will finish off this series with an answer to a question I’ve been posing for some time now: are there patterns that grow quadratically?