Tolerated Comparison, Part 1

Warning: The formula for “tolerated” comparison in this post is not correct for actual floating-point numbers. Do not use it! The next post will discuss how to compute the required value precisely in the presence of floating-point error.

Tolerant comparison is an important feature in making APL usable in the real world. But it’s also a lot slower than exact comparison. We think this tradeoff is well worth it (and of course, if you don’t want to make it, just set ⎕CT←0). But once in a while we can completely get rid of the slowdown! Added in Dyalog 17.0, along with vectorised comparison code so fast you might not even notice this change, is a cool technique that I call “tolerated comparison”. It works by turning a tolerant comparison with one argument fixed into an intolerant comparison by adjusting that fixed argument. So if you’re comparing one floating-point number to many of them, your code will run 30-40% faster than it would without this improvement.

Why tolerance?

Let’s take a trip to a slightly different APL.


A world in which tolerant comparison hasn’t been invented. Two things are equal when they’re the same.

      0.1 = 0.3-0.2
      {(0.1×⍵) = ⍵÷10} ⍳8
1 1 0 1 1 0 0 1

Okay, that’s a scary place. What happened?

The floating-point format can’t possibly represent every real number. Instead, there’s a carefully chosen set containing just shy of 2*64 numbers which can be expressed. For other numbers, we pick the closest available floating-point number. Usually this is a pretty accurate approximation. But once we perform arithmetic with a few of these values, we might go from “closest possible approximation” to “second-closest approximation” or even further away. The differences are still tiny, but when the same value—mathematically speaking—is computed in two different ways the results might be different numbers. Just barely different, but in a world without tolerant comparison that means they won’t be equal any more.

Tolerant comparison is an attempt to fix this. It’s not perfect, but it’s better in a lot of real-world situations. It makes = a useful function. The rule is that two values are equal when their difference is at most ⎕CT times the larger of their magnitudes. This is often called a “relative epsilon comparison” outside of the APL world.

Notation: To avoid confusion in the following text, I have used the APL symbols <≤=≥>≠ for intolerant comparisons, and the non-APL symbols ⋦≲≂≳⋧≉ to refer to APL’s tolerant comparisons which depend on ⎕CT. Watch out for a few quirks due to the scarcity of symbols to use: the actual APL primitives are denoted by non-APL characters, and the simpler symbol < “less-than” corresponds to the more complicated “less-than and not approximately equal to”.

Tolerant comparisons

The famous formula for tolerant equality a≂b in terms of intolerant inequality is

(|a-b) ≤ ⎕CT×(|a)⌈(|b)

This allows for a little variation in the arguments, provided that neither one is zero. And it has some other nice properties, assuming ⎕CT≤1:

  • It’s reflexive: a≂a for any number a.
  • It’s symmetric in a and b: reversing them makes no difference in the result. This is because and (|-) are symmetric functions.
  • It scales linearly: multiplying both a and b by the same nonzero value (even a negative number) does not change the result.
  • It’s convex in each variable: if a≂b and a≂d for some a, and c is between b and d, then a≂c.

In fact, tolerant equality as in APL is the only solution (or family of solutions, taking the ⎕CT parameter into account) to those constraints, other than the trivial one in which any two numbers are equal.

The above figure illustrates tolerant comparison of two real numbers. They are exactly equal only on the grey diagonal line, and tolerantly equal within the narrow strip around it. The diagram uses a much larger value for ⎕CT than Dyalog’s default of 1e¯14, or even its maximum allowed value of 2*¯32, either of which would make that strip invisible.

One way to define tolerant inequalities in terms of tolerant equality is to define, for instance, (a≲b) ≡ (a<b)∨(a≂b) using intolerant less-than and tolerant equality. But we’ll find it more useful to define and directly in terms of intolerant comparisons as with . The formula for is:

(a-b) ≤ ⎕CT × 0⌈a⌈-b

and for we can use either (a≳b) ≡ (b≲a) or (a≳b) ≡ (-a)≲(-b) to obtain

(b-a) ≤ ⎕CT × 0⌈b⌈-a

or, negating both sides and swapping the inequality,

(a-b) ≥ ⎕CT × 0⌊a⌊-b

The 0⌈ term isn’t strictly necessary (as we’ll show later), but it helps avoid some confusion when a and b have different signs. Sometimes it’s used to speed up comparison by checking if a≤b intolerantly before performing any multiplication—if it is, the multiplication isn’t needed.

Above we can see how the equation for tolerant splits into two symmetrical half-formulas, each of which is just a little off from the exact inequality a≤b. The tolerant inequality accepts either of these conditions.

The region of tolerant equality is now the intersection of the regions for tolerant and . That is:

(a≂b) ≡ (a≲b)∧(a≳b)

We’ll find this decomposition handy later on since tolerant inequalities are so much easier to work with.

Tolerating an inequality

What happens if we fix one side of the inequality? Here’s a diagram for a≲b with a dashed line drawn where b is equal to some fixed B and a varies.

There’s only one place where the black line bounding the region where a≲b intersects the line b=B. The value of a at this intersection has an interesting property: when a is (exactly) less than or equal to that value, then it is tolerantly less than or equal to B. If we want to compare B to many other floating-point values, and we know the a coordinate of the intersection above, then we can skip all of the tolerant comparison arithmetic and just use exact comparison with that value!

Once we have a way to do this for all other inequalities will follow, since (a≳B) ≡ (-a)≲(-B) and (a⋧B) ≡ ~(a≲B).

Maybe you can figure out the value from the graphs above. But it’s also easy to find the value (and prove it works!) using a fact about the maximum function: for real numbers x, y, and z, x≤y⌈z is the same as (x≤y) ∨ (x≤z). One of the advantages of treating inequalities as functions like any other is that formulas like these are easy to manipulate. Here we go:

Start with the formula for tolerant inequality, writing q in place of ⎕CT.

(a-B) ≤ q × 0⌈a⌈-B

Distributing over , we have

(a-B) ≤ 0 ⌈ (q×a) ⌈ (-q×B)

and using (x≤y⌈z) ≡ ((x≤y) ∨ (x≤z)) twice (just like in the previous step, this can be thought of as distributing x≤ over three arguments of an associative function all at once), we can expand this formula to

((a-B) ≤ 0) ∨ ((a-B) ≤ q×a) ∨ ((a-B) ≤ -q×B)

which simplifies to

(a ≤ B) ∨ ((a×1-q) ≤ B) ∨ (a ≤ B×1-q)

or (here we need q<1, so that 1-q is positive; otherwise we’d have to reverse the middle inequality)

(a ≤ B) ∨ (a ≤ B÷1-q) ∨ (a ≤ B×1-q)

Recombining, we get

a  ≤  B ⌈ (B÷1-q) ⌈ (B×1-q)

So that’s our formula: now the tolerant equality is an intolerant equality, where the right side depends only on B!

Reading the signs

So we have a formula for a “tolerated” value b = B ⌈ (B÷1-q) ⌈ (B×1-q). When do the different possible maxima come into play?

  • If B=0, then everything on the right is zero, and b=0.
  • If B>0, then B×1-q is smaller than B, since q>0. But that means ÷1-q is larger than 1, and B÷1-q is larger than B. So b=B÷1-q.
  • If B<0, then the all the inequalities that we got with B>0 are reversed. This means b=B×1-q.

Note that none of these cases depend on the branch b=B, even though the first one touches it. That means the definition of tolerant inequality would be the same even if we remove that branch. By working backwards through the above derivation we find that the initial 0⌈ has no effect, and we can remove it from the definition.


Tolerant equality obviously can’t be reduced to intolerant equality, since many values can be tolerantly equal to a number but only one is intolerantly equal to it. But recalling that (a≂b) ≡ (a≲b)∧(a≳b), we can express it as two intolerant inequalities: tolerate b for as above. Then tolerate it for (this always yields a value at least as large as the first). a is tolerantly equal to b if it lies between the two tolerated values.

This case is particularly important for the set functions ⍳∊∩∪~, which by definition use tolerant equality to match values in one argument to values in the other. If the non-principal argument (the one containing values to be looked up, like the right argument to ) is short enough, then each element in it is looked up using a simple linear search through the principal argument. This means that one number at a time is compared to the entire principal argument (up to the first match found). So for floats we can get a substantial improvement by tolerating each element on the right before comparison.


That was the easy part.

In this post we’ve been assuming that floating-point numbers behave like mathematical real numbers. Given that the whole reason tolerant comparison was introduced is that they don’t, maybe that wasn’t the best assumption.

The formulas B÷1-q and B×1-q for tolerated comparison are flawed: they produce values that differ slightly from the actual number that should be used in an intolerant equivalent to tolerant comparison. This is very dangerous: many programs could break if two values compare equal in one comparison but not in another! In the next post we’ll dive into the world of floating-point error analysis and find out how to get the right values. And we’ll discover that imperfect computation isn’t all bad when we find a faster alternative to a very slow floating-point division.

Expanding Bits in Shrinking Time

The chart above compares the performance of ∘.∧ in Dyalog versions 16.0 and 17.0. The improvement is a factor of 3-4 across the whole range (except at multiple-of-8 lengths, where 16.0 spikes up). That’s not easy to get, especially for a function like ∘.∧ which was already the target of a fair amount of special code. And the improvements are completely general. They apply to any scalar dyadic which has a Boolean result on Booleans, which is all but +-○.

How did it happen? The answer, at least for the parts of the graph left of 1024, follows. It takes us through a surprising number of APL primitives, which I think is a great demonstration of the ways APL thinking can lead to faster algorithms even in other languages.

Binary choices

One of the many performance bumps in Dyalog version 17 is that selecting from an array of two options using a Boolean array is much faster. Dyalog 17 also takes advantage of this speed improvement by using it for a few other functions: scalar dyadics when one argument is a singleton and the other is Boolean, (or ) where the right (left) argument is Boolean, and outer products where the left argument is Boolean.

Selection has been sped up for all kinds of indexed array, but the most interesting case is that in which the cells of the indexed array are Boolean, with a cell shape that’s not a multiple of 8 (when working with Boolean algorithms, we tend to call such shapes “odd”). In this case the rows of the result aren’t byte-aligned, so to generate them quickly we will need some Boolean trickery. As it turns out, a lot of Boolean trickery.

The centre of our attention will be the function Replicate (/) when the right argument is Boolean and the left is a scalar. That means something like 3/1 0 0 1 0. Replicate doesn’t seem obviously connected to indexing, but like indexing, it turns each bit of the argument into multiple bits in the result. In fact, this is by far the most important and difficult step in implementing Index, since Index may be phrased straightforwardly in terms of Replicate using exclusive or ():

      y[x;]  ←→  (((≢y)/⍪x) ∧⍤1 ≠⌿y) ≠⍤1 ⊣⌿y

If for some reason that doesn’t look straightforward to you, I’ll explain how it all fits together later. But first, Replicate!

The trouble with Replicate

Just implementing Replicate on a Boolean array isn’t difficult. Read one bit at a time from the argument, and then write it to the result the required number of times. However, reading and writing one bit at a time is slow. A fast algorithm for Replicate will work an entire machine word at a time (that is, 64 bits on 64-bit systems).

But even fairly well-implemented word-at-a-time strategies will quickly run into another problem: branching. A simple if/then construct in C can be very cheap or very expensive depending on whether the CPU is able to predict in advance which path will be taken. With odd-sized replicates the not-quite-periodic way that the writes interact with byte boundaries will make nearly all useful branching unpredictable. But there are many techniques which can allow the CPU to make choices without having to change the path it takes through a program. Coming up is one rather exotic entry in the field of branchless programming.

Small expansion factors

When beginning work on fixed-size replicates, I quickly found that, in Dyalog version 16, it was much faster to add a length-1 axis before the first, replicate that, and transpose it to the end than to replicate the last axis directly.

      ⎕IO←0 ⋄ x←?1e4⍴2
      cmpx '5/⍪x' 'x,,⍨,[0.5]⍨x' '⍉5⌿(1,⍴x)⍴x'
  5/⍪x         → 9.5E¯5 |   0% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕
  x,,⍨,[0.5]⍨x → 3.4E¯5 | -64% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕
  ⍉5⌿(1,⍴x)⍴x  → 1.3E¯5 | -86% ⎕⎕⎕⎕⎕⎕

This is largely because I had spent a lot of time improving Transpose with Boolean arguments of various shapes in version 16 (the comparison above looks very different in version 15). Transpose on a short but wide Boolean matrix uses a technique that I call bit interleaving, which would unfortunately require another blog post to explain. To transpose a matrix with r rows, I move across the rows, using some special tricks to insert (r-1) zeroes after each bit in each one, and combine them using bitwise or. To instead replicate a vector by r, I modified this to expand the one argument as though it were going to be combined with other rows, but instead of doing that, I combined it with itself by multiplying by the number whose base-2 representation consists of r ones. (C programmers may notice that multiplying by this number is the same as shifting to the left by r and then subtracting the original value. In fact, using multiplication is faster—x86 variable shifts are slow!)

Bit interleaving is a somewhat complicated algorithm, but it had all been done by the time I started on Replicate, and modifying it didn’t take long.

Large expansion factors, the obvious way

The bit-interleaving algorithm is very fast for small numbers and gets slower as the expansion factor increases. Above 64 it is no longer usable because it only writes one word at a time, while each bit should expand to more than a word. I chose to stop using it after 32, as it turns out we can get just about the same performance at 33 using a different approach.

The old approach, on the other hand, is only about half as fast at this expansion factor. Version 16’s algorithm reads from the right argument one bit at a time, then writes that bit the appropriate number of times to the result by writing one partial byte and then using memset (which may spill over into the next bit’s area since memset can only write whole bytes; it will be overwritten when we get to the next bit). This works fine for large enough left arguments, and scales up very well: memset is provided by the C standard library and will push bytes to memory as fast as possible. But for small values it is not so good: the call to memset branches a few times based on the unpredictable size of the write, which incurs a steep branch misprediction penalty.

There’s no easy way to get around the penalty for writing some bits at a variable offset. If the expansion factor isn’t a multiple of eight, then the number of bytes each expanded bit touches will always vary by one, and having to check whether to write that byte results in an unpredictable branch. There are ways to avoid a check like this (overwriting is one), but these tend to be fast only on a small range of expansion factors and to be very tricky to write and bug-prone, especially at the boundaries of the result.

A new idea

So the problem is that writing a weird number of bits at a time is expensive. But it’s possible to get away with only writing one bit for each “memset” we want to perform! The trick is found in a pair of inverse functions well-known for their usefulness in Boolean APL code. These are {2≠/0,⍵} and ≠\, and they are analogous to the pairwise difference {¯2-/0,⍵} and progressive sum +\. Dyalog 16 has fast code for ≠\ provided by Bob Bernecky, so I knew it was a good building block for other high-performance primitives.

The diagram above illustrates the relationships between scans and pairwise differences. In the bitwise world consisting only of even and odd, is like +. But each number is its own negative, so it’s also like -. More precisely, since ⍺≠⍵ is the same as 2|⍺+⍵ on Booleans (check for yourself!) and modulus distributes over addition when the final result has a modulus applied to it, ≠\ is equivalent to 2|+\. Similarly, {2≠/⍵} or {¯2≠/⍵} is the same as {2|¯2-/⍵}. The diagram includes the extra zero in {2≠/0,⍵} as part of the vectors in the top half of the diagram, but omits it from the functions used to label arrows to avoid clutter.

Let’s take a look at the pairwise difference after we replicate a vector:
It’s very sparse—the only one bits come at indices which are multiples of five. This is because expanding v multiplies the indices of its pairwise difference by five:

      v←1 1 0 1 0 0 0 1
      ⍸ 2≠/0, v  ⍝ ⍸⍵  ←→  ⍵/⍳⍴⍵
0 2 3 4 7
      ⍸ 2≠/0, 5/v
0 10 15 20 35

We can recover 5/v using the inverse ≠\ of pairwise difference. So to expand, we can add four zeros after each bit under the ≠\⍣¯1 function (which is the same as pairwise difference):

      ≠\ ,5↑[1]⍪ ≠\⍣¯1⊢ v
1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1

Of course isn’t particularly fast on these arguments, so we will have to write our own version, which sets the entire result to zero and then writes bits in the appropriate places. Implemented properly, this goes pretty quick, the only substantial costs being a variable shift and one byte-sized write per element on the right. It’s completely branchless. Follow up with the fast ≠\ algorithm and we have a pretty good solution.

This solution performs more writes than it has to, since it writes each difference bit even if it is zero. But trying to get rid of the extra writes is a losing battle. Checking the value before each write takes much longer than just writing it, and even a fancy strategy, like using the count-leading-zeros instruction to quickly skip over all zeros before each one, will be slower for arguments without many zeros in the difference vector. But there is another way to make this algorithm a little faster, coming up after we tie up some loose ends…

Back to Index, and more xor

What about my unexplained formula for indexing? The expression

      y[x;]  ←→  (((≢y)/⍪x) ∧⍤1 ≠⌿y) ≠⍤1 ⊣⌿y

also relies on insight about , but in a completely different context. Let’s simplify it a bit to see what’s going on. Suppose we have two Boolean arrays b and c of the same shape, and we want to use the bit a to select one of them: we want an expression that returns b if a=0 and c if a=1. For real numbers, we might use a linear combination:

      b + a×(c-b)

If a=0, then the c-b term is removed and we get b. If a=1, then it stays and we have (b+(c-b)) = c.
This formula is also valid for Boolean arrays, of course, but introducing an integer c-b would be a poor choice from a performance standpoint. Instead, we can use the insight that performs the function of both + and - in the bitwise world, and transliterate our formula into

      b ≠ a∧(c≠b)

Again, if a=0, then the c≠b is zeroed out, and we end up with b≠0, which is the same as b. If a=1, then it stays, leaving (b≠(c≠b)), which gives c.

If y is a shape 2 m array y←b,[¯0.5]c, then we have b ≡ ⊣⌿y and (c≠b) ≡ ≠⌿y. So b ≠ a∧(c≠b) is, rearranging some, the same as (a ∧ ≠⌿y) ≠ ⊣⌿y. But there is an implicit scalar extension in the function; we can make this explicit by writing ((m/a) ∧ ≠⌿y) ≠ ⊣⌿y. Now it’s clear that to select using each bit of a vector x, we need to turn each bit into a row with length m, and then apply the and functions on rows, giving the formula I showed earlier.

How about Outer Product? That one’s easier, with no Boolean magic required. For vector arguments, ⍺ ∘.f ⍵ is identical to ((≢⍵)/⍪⍺) f ((≢⍺),≢⍵)⍴⍵, and the same computation works for general arrays if we use the ravel length ×/∘⍴ instead of the count . When the right argument has more than around 1024 bits, replicate becomes slower than the old way of just pairing each bit of the left argument with the whole of the right, so we use that method instead.

Speeding up ≠\

We left off with an expansion algorithm whose cost is dominated by the time to compute ≠\ on a vector. That computation is based on a fast method for ≠\ on a single machine word. It moves along the argument vector (which is the same as the result vector in this case) one word at a time, keeping track of the most recent bit of the result. Then each word of the result is obtained from the fast xor-scan of that word in the argument, xor-ed with the carried bit. For arbitrary input vectors, this is the best algorithm. But the word-length ≠\ computation is slowing us down, and it turns out we can do a lot better.

Writing a single bit has the same cost as writing an entire aligned machine word. So writing one bit at a time, while much better than writing a variable number of bits with arbitrary alignment, represents some wasted effort. Can we make the impending ≠\ computation easier by using full-word writes?

A definite “yes” for that question. In fact, we can eliminate all of the intra-word computation. Consider the effect of a bit that we write on the final result after applying ≠\. That bit will be spread out over all of the subsequent bits, flipping each one. In order to produce this effect on a single word (leaving the task of flipping subsequent words for later), we can thus xor an aligned word with a word which is zero before that bit and one at that bit and later. We can get this word just by shifting the all-ones word right by an appropriate amount. And then we can omit the expensive word-size ≠\ computation entirely, so that to perform the final pass we simply repeatedly xor the next word with the current one’s last bit. This greatly reduces the cost of the final pass at very little expense to the pass before it (xor-ing rather than setting a word requires reading that word’s value first, so it’s not quite free).

To summarize, there are three passes:

  • Set all bits of the result to zero.
  • For each bit in the right argument, xor the appropriate word in the result with a word which is all ones after the place where that bit starts in the result.
  • Iterate across words of the result, xor-ing each with the last bit of the previous word (after finishing computation on that word).

For expansion factors between 32 and 256, this is the fastest method I know of. Below 32, interleaving is faster, and above 256 simply using memset will do the trick.

The xor-based algorithm also works for Replicate or Expand when the left argument is a vector, and in Dyalog 17 it is used for both of these whenever the average expansion factor, which is equal to the ratio of result length to argument length, is 256 or less.

The end result

The graph above shows the gains in scalar-Boolean replicate between versions 16.0 and 17.0. Note the huge scale of the vertical axis—replicate is just short of a hundred times faster at the far left! Version 17’s Replicate is split into three parts, with the performance of each individual algorithm shown under the overall result. From left to right, these are the algorithm with bit interleaving, the xor-based algorithm described in the last section, and an algorithm with memset. The last of these is the same basic strategy used by version 16, but there are some improvements in version 17 that make it about 45% faster for short arguments.

The seams between these algorithms are not quite perfect on my machine, but they are not off by enough to damage performance significantly. The cutoffs favor the middle, xor-based algorithm a little. I think this is appropriate because that algorithm is likely to have consistent performance even on older machines and different architectures. In contrast, bit interleaving uses the BMI2 instruction set (available on x86 processors since 2013) and is slower if it is not present, and memset may perform differently with other implementations of the standard library or depending on available vector instruction sets. Our xor-based algorithm uses only very widely available instructions, and one of the costs—the variable shift used on the all-ones word—is actually much faster on other architectures because of a design flaw in x86. An excellent addition to Dyalog APL!

Stackless Traversal

Enlist (∊) is twice as fast in Dyalog 16.0 as it was in Dyalog 15.0. Pretty much across the board: ∊⍳100 is not going to be any faster, but whenever the argument is a nested array and the simple arrays it contains are reasonably small, there are huge performance improvements. How did we achieve the huge speedup?


The usual way for a C programmer to write the traversal used in Enlist would be a simple recursive function: If the current array is simple, handle it, and if it is nested, traverse on each array it contains. We can’t do this, though, because it would break our product. The specific problem is that the C stack used for function recursion has a limited (and fairly small) size, while the depth of nesting in an array is limited only by available memory. When passed a tree a few thousand layers deep, an Enlist implemented using the C stack would run out of space, then attempt to write past the end of the stack, resulting in a segfault and a syserror 999.
So keeping memory on the stack is out. The only place we can store an arbitrary amount of memory is the workspace itself. But storing things in the workspace is harder than it sounds. When Dyalog fails to do a memory allocation, it tries to get more space by compacting pockets of memory and squeezing arrays. This happens rarely, but any operation that allocates memory has to take it into account. That means it has to register all of the memory it uses so it can find it again when memory is shuffled around.

The old approach: trav()

Dyalog has a general function for traversing arrays, which will take care of all of the problems above. It’s very flexible, but in enlist it’s used for a single purpose: to call a function on each simple array (“leaf”) in a nested array. The old Enlist used it twice, first to calculate the type and length of the final result, and again to move all of the elements into a result array after it’s allocated.
Trav works very well, but is also very slow. It stores a lot of information, and constantly allocates memory to do so. It also is much more general than it needs to be for enlist, so it spends a lot of time checking for options we never use. We can do better.

A different strategy

Rather than allocating a bunch of APL pockets to emulate the C stack, what if we just didn’t use the stack at all? Sounds kind of difficult, since we definitely need some sort of stack to remember where we are during traversal. Once we finish counting or copying the last leaf in a branch, we have to find our way back up to the pocket we started at. This pocket could be anywhere from one to a billion levels above us, and we have to remember the location of every pocket in between. And once we return to a pocket, we have to remember how many children we have already traversed, so we can move on to the next one. Where do we store all of this information?

Fun with pointers

Suppose we are in the middle of traversing a nested array p. An array in Dyalog is a pointer to a pocket of memory, which contains a header followed by some other pocket pointers p[0], p[1], and so on. We consider the pocket pointers themselves, rather than the data that they point to, to be arrays since pointers can be stored and manipulated directly and pockets cannot. If the rank of p is more than one, p[i] refers to the i’th array in ,p.
Let’s assume we have gone through the first array p[0] in p and now want to begin traversing p[1], which we rename q. In order to move on to p[2] after we finish with q we will need some stored information—a stack-based algorithm might push p and the current index 1 onto the stack. A variable-size stack is needed rather than some constant amount of memory because this information has to be stored at every level of the arbitrarily-deep traversal. We don’t want to use such a stack, but there is a redundant value we can use: the pointer p[1]=q stored in p. When we finish traversing q and come back to p, we know the address of q, since we just traversed it! So we can safely overwrite that address. But all of the other nearby pointers are in use. It would be very difficult to find space for another word. So instead of storing the location of p, we store a slightly different address: ptr, which points to p[1] (we’ll discuss how to recover the pointer p from ptr later). It won’t help to store ptr in p, since it would just point to its own location! Instead we use a temporary variable prev to hold onto it while we traverse q, and store the previous value of prev, which points to a location inside p’s parent, in the location that used to hold q. By the time we begin traversing q’s first child, r, the picture looks like this:
We have reversed all of the pointers above us, and now we have a trail leading all the way back up to the array we started at. To do this we needed an extra temporary value, prev. This uses more memory, but only a constant amount (one word), so it won’t cause problems for us. However, it might be helpful to consider why we need that extra word. In a stack-based traversal only one temporary pointer would be needed to iterate over the array, and in fact we do get away with only writing one word of data in each array during traversal. But the first array doesn’t have any parent array, so there is no pointer to write in it. Instead we write a zero (which is not a valid pointer) to indicate that fact. This zero is an extra word that doesn’t correspond to any array, so to compensate for it we use an extra temporary variable.

When to stop

When we finish traversing a pocket, we end up with another problem. We don’t know we’re done! Following the algorithm as described so far, we would end up with ptr pointing at the last element of q, which looks like all of the others, and we would keep moving past the end of q, landing on the header of the next pocket, which probably isn’t even a pointer. Segfault—do not pass go; do not collect the elements of the original array into a list.

A typical algorithm would just store the length of the array on the stack, which we can’t do. However, there is a little more space hiding in the array q which we can use. A pointer gives the location of a particular byte in memory. But the pointers in q all point to the beginning of pockets, which are word-aligned: they begin at addresses which are all multiples of 4 bytes (in 32-bit builds) or 8 bytes (64-bit builds). So we have at least two bits at the bottom of each pointer which are guaranteed to be zero. As long as we remember to clear those bits once we finish, and before following a pointer, we can write anything we want in them!
The first thing to write is a stop bit. We mark the last pointer q with some two bits which aren’t zero (we’ll decide what later). While shuffling pointers around, we always leave the bottom two bits where they were, so that when we get back to q[n] we see the two bits we wrote when we started traversing q. If we haven’t reached the end, that could be anything other than the stop code, and if we have reached the end, it must be the stop code. So now we know when to stop, and won’t segfault by running past the end of a pocket.
However, in order to get back to the parent array with all of our state intact (specifically, the overwritten pointer to the current array), we need to know the address of the beginning of our pocket, not the end. And the shape of the pocket, which we would need to find the pocket’s length, is kept at the beginning of the array as well. How do we get back to the start?

Retracing our steps

Obviously we need to start moving backwards. But once again, we need to know when to stop. Since we’re about to run into it face-first, maybe we should discuss what comes before the pointers in a pocket?

The contents of an array pocket are pretty simple. There are three words at the beginning: the length and reference count of the pocket (which we don’t need to worry about), and the zones field, which contains the type and rank of the pocket along with potentially some other data. Next is the shape (which might be empty!), and then the data in the array—for a nested pocket, pointers to other pockets. There are three cases we want to consider:

  1. The shape is empty, which means q contains only one pointer. We will run into the zones field right before the last (and first) pointer.
  2. The shape is not empty, but the total number of pointers is small. We will run into the shape after a short amount of backwards searching.
  3. The total number of pointers is large. It would take a while to get back to the beginning one pointer at a time, but we also have a lot of spare bits at the bottom of those pointers…
  4. (Extra credit) The pocket is empty, but it still has a prototype to traverse. For enlist we ignore these pockets, but for other purposes we have to handle this case. The shape has to have a zero in it, but the other axes could have any size that fits in a word.

For the first two cases, we mark the shape and the zones field, respectively. By a lucky coincidence, the bottom two bits of the zones field are used to store the type of the pocket we are in, and for nested arrays, the two bits we care about are always both 1. So let’s make 3 (in binary, 11) the indicator for the zones field. In case 2, the last item of the shape is small, so there are free bits at the top. We’ll shift the shape up to move those to the bottom, leaving room to store the rank and a marker. We’ll choose 1 for the marker.
In case 3, there are enough pointers to store the length of the array in the bottom bits. Since we can’t use the stop marker to encode the length, we opt to just use the bottom bit, that is, storing zeros and ones in the bottom two bits, of each pointer. To distinguish from case 2 we mark the beginning of this offset with a 2, the only unused code so far. We mark the end of the offset with a 2 as well, and then to read the offset we just move backwards, adding each bit to the end of an integer, until we reach that 2. Then we use the length to get back to the beginning of an array. While moving backwards, we also clear those bits to leave the pointers pointing at the exact, byte aligned start of pockets like they did when we found them.
Case 4 is best left as an exercise, I think. There’s a whole empty word in the shape for you to use—what could be easier? Granted, it’s stashed away in some random location in the shape, but that shouldn’t be an obstacle.

Conditions of use

The traversal described above is much faster than trav, but it requires more care to use. Since it scribbles notes on parts of arrays during traversal, the workspace isn’t necessarily in a valid state until it finishes. So it’s not safe to call memory management functions during traversal. Any memory used needs to be allocated before the start of traversal, which means it has to have a constant size: we can’t use anything that will grow as needed.
Enlist is made up of two very simple functions which meet this requirements in most cases: the first goes through and records what types are in the array and how many total elements there are, and the second actually moves the elements into a result array. For the counting function, we just need a little extra storage to fit the type and the count. That’s a constant amount, so it’s fine. For the moving function, we can’t promote an element of a non-mixed array to an element of a mixed array, since this requires us to allocate a new array for it. But this is not a very common case, and when it does happen it will be slow anyway. This means it’s okay to use trav for that case, and use the faster traversal otherwise.


Here are timings showing how long it takes to enlist various arrays, and the improvement from version 15.0 to version 16.0. Timings are measured on a modern CPU, an Intel Kaby Lake i7, but I don’t expect the ratios to change very much on other machines.

APL expression Description 15.0 16.0 Ratio
{(?⍵⍴10){⊂⍵}⌸?⍵⍴99} 1e3 1e3 bytes in 10 groups 6.138E¯7 2.898E¯7 2.118
{(?⍵⍴10){⊂⍵}⌸?⍵⍴99} 1e4 1e4 bytes in 10 groups 1.174E¯6 9.439E¯7 1.244
{(?⍵⍴10){⊂⍵}⌸?⍵⍴99} 1e5 1e5 bytes in 10 groups 8.799E¯6 8.168E¯6 1.077
{⍵⊂⍨1,1↓1=?(≢⍵)⍴2}⍣10?1e4⍴99 1e4 bytes nested 10 deep 8.540E¯4 3.464E¯4 2.466
⊂⍣1e3⊢2 3 A 1000-deep nesting 1.032E¯4 1.533E¯5 6.730
,⍨∘⊂⍣20⊢2 3 Duplicated references 1.648E¯1 3.595E¯2 4.584

The first row is probably the most typical case: a single array with a little bit of nesting, but fairly small leaves. In this case, we can expect to see about a factor of two improvement. As the leaves grow larger, the running time becomes dominated by the time to actually move data to the result, and the improvement falls off, dropping to 25% around 1000 bytes per leaf and 0.8% at 1e4 bytes per leaf. The new code is faster even at very small numbers of leaves. In fact, it takes less time to start up an array traversal than the old method, so it’s many times faster on very deeply nested arrays with few elements in each array. We can see from the final row that the new method has no issues when it encounters the same array many times in one traversal.
The same algorithm is also used for hashing arrays in Dyalog 16.0, which in turn is used for dyadic iota and related functions. The impact on individual operations is not quite as much as the improvements to enlist shown above, but they are still substantial, and being able to search for a few strings in a list more quickly will surely help a lot of applications.