I was recently writing some performancesensitive code in which I had a double
array of distances (one per element), and I wanted to get a list of elements
sorted by distance:
Java provides Arrays.sort
for direct sorting; that is, it’s easy to ask it to
sort distances
or 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:
Element
s each in constant timeI 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:
Now distances[]
is sorted such that the distance of
elements[(int) (Double.doubleToLongBits(distances[i]) & ~mask)]
is ascending
for ascending values of i
.
Each distance
is encoded as a doubleprecision float, which internally looks
like this:
+ 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 loworder 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 a
and b
and we lose
24 bits:
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 loworder mantissa bits to store information. In this case I’m storing the original array index for each distance in its loworder bits. Here’s the code above, piece by piece:
Then we tag each distance this way (here I’m assuming we’ve got between 2^{19} and 2^{20} elements, so we reserve 20 bits):
d[i] S EEEEEEEEE MMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM
original bitstag 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
double
s. Even a welloptimized 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)
elementcopy operations.
It is possible to simulate bitwise access using floating point arithmetic and int casting (which flotsam does in Javascript), but it requires some care – particularly in this case, when any rounding error will cause data loss within the indexes. It’s also a lot slower than the bitwise solution above, possibly enough to outweigh any performance benefits in the sorting logic itself.
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 partspertrillion error we’re introducing doesn’t really pose a problem most of the time). Not all distributions are so robust to bittwiddling, 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.