I was recently writing some performance-sensitive code in which I had a
array of distances (one per element), and I wanted to get a list of elements
sorted by distance:
Arrays.sort for direct sorting; that is, it’s easy to ask it to
elements by its natural ordering. But in this situation,
the two arrays are tied together only by indexes, which would require a
comparator to maintain a reverse lookup from
Element to either its index, or to
its distance. That’s a lot of overhead – particularly because the map would
require generic boxing of either type of value.
Luckily there’s an interesting way to solve this problem that meets the following requirements:
Elements each in constant time
I should also mention that the resulting ordering isn’t exact, but it is very close.
Before I go into how I solved it (which took a while to think of), you should see what you come up with. It’s a fun problem.
First in code:
distances is sorted such that the distance of
elements[(int) (Double.doubleToLongBits(distances[i]) & ~mask)] is ascending
for ascending values of
distance is encoded as a double-precision float, which internally looks
+--- sign | | +- exponent | | S EEEEEEEEE MMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM |---------------------52 mantissa bits-------------------|
The mantissa for all but the smallest numbers is normal, meaning that it’s
interpreted as though there were a leading
1 in a 53rd bit. This puts an upper
bound on the significance of low-order mantissa bits, which is what we need for
the code above to work.
Depending on how the distances are distributed, we can make a probabilistic
argument about how much the ordering will change as we lose precision in the
mantissa; specifically, suppose we’ve got two distances
b and we lose
A = S EEEEEEEEE MMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM B = S EEEEEEEEE MMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM |-----------------------------| |------------------------| keeping these bits losing these
The probability of changing the ordering between these two points is the same as
the probability that the bits we’re keeping are all identical between them.
Benford’s law converges rapidly
to a uniform distribution for subsequent digits and the leading
1 is implied,
so in practical terms P(reordering) is very nearly 2-k, where k is
the number of bits being kept.
If we can lose some precision without causing problems (which for my use case was true), then we can arbitrarily reassign low-order mantissa bits to store information. In this case I’m storing the original array index for each distance in its low-order bits. Here’s the code above, piece by piece:
Then we tag each distance this way (here I’m assuming we’ve got between 219 and 220 elements, so we reserve 20 bits):
d[i] S EEEEEEEEE MMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM |-----------------original bits----------------||-----tag space------| & mask 1 111111111 1111 11111111 11111111 11111111 11110000 00000000 00000000 | i 0 000000000 0000 00000000 00000000 00000000 0000IIII IIIIIIII IIIIIIII = S EEEEEEEEE MMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMIIII IIIIIIII IIIIIIII
At this point
Arrays.sort() will be none the wiser and will sort the array
normally (and, importantly, very quickly).
Now we can read the tags back to recover the ordering, which looks like this:
mask 1 111111111 1111 11111111 11111111 11111111 11110000 00000000 00000000 ~mask 0 000000000 0000 00000000 00000000 00000000 00001111 11111111 11111111 d[i] S EEEEEEEEE MMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMIIII IIIIIIII IIIIIIII & ~mask 0 000000000 0000 00000000 00000000 00000000 0000IIII IIIIIIII IIIIIIII (int) 00000000 0000IIII IIIIIIII IIIIIIII
You can use this hack in any language to similar effect, though you lose most
of the performance advantages if the language doesn’t have bitwise access to
doubles. Even a well-optimized sorting function that doesn’t have the Java
indirection problem will benefit if you can store the data in the floats
directly, since the array will be smaller in memory and you’re doing O(n log n)
Distances are ideal for this type of hack because they tend to be spread over a wide range of magnitudes, and even when they aren’t, you don’t tend to care much whether the points are exactly ordered (i.e. the parts-per-trillion error we’re introducing doesn’t really pose a problem most of the time). Not all distributions are so robust to bit-twiddling, though. In particular, if you had a case where the variance were many orders of magnitude smaller than the average – e.g. 1,000,000,000 ±0.003 – then there’s a good chance the small bits would matter. It’s important to figure out the probability of a false reordering before losing bits of precision.