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 2^n-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 2^n-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 2^k 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 13^n-3 base case calls in order to step forwards 2^n-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?

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 9^n-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?