8 unstable releases (3 breaking)
✓ Uses Rust 2018 edition
|0.4.2||Apr 29, 2019|
|0.4.1||Apr 29, 2019|
#304 in Algorithms
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Hamming Weight Tree from the paper Online Nearest Neighbor Search in Hamming Space
To understand how the data structure works, please see the docs.
Most recent benchmark for 1-NN:
You can find benchmark output here.
If you would like to run the benchmarks yourself, just run
cargo bench at the command line. I recommend using
RUSTFLAGS='-C target-cpu=native' cargo bench instead since both linear search and this tree are both significantly faster when using modern instructions.
The Hamming Weight Tree was originally implemented in the paper "Online Nearest Neighbor Search in Hamming Space" by Sepehr Eghbali, Hassan Ashtiani, and Ladan Tahvildari. This is an attempt to improve on the performance and encapsulate the implementation in a Rust crate for easy consumption.
Here is how we would like to think about a number visually, in terms of a binary tree of its substring hamming weights:
5 3 2 2 1 1 1 1 1 0 1 1 0 0 1
B be the log2 of the width of the number. In this case
B = 3,
2^3 = 8.
L be the level of the hamming tree. The hamming
weight of the whole number is the root and is
L = 0.
N be the index at the level
L of the substring weight in question.
W be a weight of the substring
N at level
MAX be the side max of the hamming tree.
MAX = min(W, 2^(B - L - 1)).
This is the maximum number of ones that either side of a substring can
MIN be the side min of the hamming tree.
MIN = W - MAX.
This is the minimum number of ones that either side of a substring can
Every time we encounter a weight
W in the tree then the next two
substrings can vary from
[MIN, MAX] to
[MAX, MIN] for a total of
A + 1 possibilities. Therefore we can also view the tree like this:
5 [1-4] 2 3 2 [1-2] [0-2] 1 1 2 1 1 1 [1-1] [0-1] [0-1] [0-1] 0 0 1 0 1 1 0 1 1 0 0 1
On the left we have the actual tree. In the middle we have the
possible values for the left branch. On the right we have the index
of the left branch chosen, which is calculated by subtracting the left
substring weight by
To compute the index for
L we must iteratively multiply an accumulator
MAX - MIN + 1 of the current substring
N, add the substring's index
from the tree, then shift the number over by the substring width to get
N + 1.
To do the reverse, we must mod the accumulator by the multiplication of
all lower substring
MAX - MIN + 1 to get the index of that substring
and then divide by the
MAX - MIN + 1 of the current substring.
Do this iteratively to produce all weights for a given index.
We should avoid computing the weights from the index more than once
per operation if possible because it is costly due to modulo and division.
To limit the search space, we depend on the fact that the sum of the
absolute differences of hamming weights of substrings cannot exceed
the sum of the hamming distances of substrings. This means if the
sum of the absolute differences in hamming weights between the
bucket index's implicit weights at any given level of the tree
radius then we know it is impossible for any results to be
found in that branch of the tree. This allows us to filter what we
search to be only nodes that could theoretically match.
For the top level, its clear to scan (
This is because results cannot be found outside where the weight differs
by more than
radius. For the levels below that it becomes more
complicated to search the bucket. To do so, let us consider the case of
L = 0 (the 0th level starts after looking up the bucket for the overall
Lets say we have a 128-bit feature with this tree of hamming weights:
5 3 2
If we want to search for things in
radius <= 1 then at the top level we
4..6. Let us consider what happens when we then try to search the
bucket found at index
4. At this point we have a situation where the left
side could vary in
0..=4, since we have a 128-bit number, each half can
4 ones. However, we dont need to search all of these
If the left side were to have a weight of
1 then the right
side would have a weight of
3. Remember "the sum of the absolute
differences of hamming weights of substrings cannot exceed the sum of
the hamming distances of substrings." If we look at our search point, we
find that the sum of the differences is
abs(3 - 1) + abs(2 - 3) = 3.
This is greater than our search radius of
1, therefore it is impossible
to find a number with a hamming distance within the radius there.
Now consider what happens if we go to a weight of
2 on the left side.
In this case we have
2 bits on the right side. The sum of the differences
abs(3 - 2) + abs(2 - 2) = 1. This is equal to our search radius and
therefore it is possible to find a match in that bucket.
In conclusion, we need to iterate in
2..=3. This has limited the
possibilities greatly. However, we need to know how to derive this range.
What we are going to find specifically is the way to derive the range of
the left substring weight (not the actual bucket index) that allows just
that substring to fit inside of a
radius. We will use this primitive to
derive the solution for any number of substrings.
Let the weight of the target parent substring be
Let the weight of the target left substring be
Let the weight of the target right substring be
Let the weight of the search parent substring be
Let the weight of the search left substring be
Let the weight of the search right substring be
TR = TW - TL
SR = SW - SL
Let the sum of substring weight differences be
SOD = abs(TL - SL) + abs(TR - SR)
We are searching for
TL that satisfy
SOD <= radius. The
SOD has two
inflection points that come from the two
abs in its expression. Between
those two inflection points there are only four possible combinations:
TLis going towards
TRis going towards
TLhits its its inflection point first and starts going away from
TRis still going towards
TRhits its its inflection point first and starts going away from
TLis still going towards
TRhave both hit their inflection points and are going away from
As we can see, regardless of whether
TR hit their inflection
point first, we can be guaranteed that the slope is
0 before the final
inflection point. This happens because
TR are inversely related.
We must start by computing where the first slope would intersect with the
radius. We assume that
TL is below or equal to
SL and that
above or equal to
SR. Given this, we know that when
(SL - TL) + (TR - SR) = radius we enter the search area. Since
TR = TW - TL we can rewrite this as
(SL - TL) + (TW - TL - SR) = radius. Since
all known at this point, we can solve for
TL = (-radius + SL - SR + TW) / 2
Lets do the same thing for the opposite case (slope
(TL - SL) + (SR - TR) = radius
(TL - SL) + (SR - TW + TL) = radius
TL = (radius + SL - SR + TW) / 2
We can see that there is a shared intercept between the two equations, but we will not extract the intercept directly because we wouldnt get the same result if we divide by 2 before adding since we would loose a bit of precision.
C = SL - SR + TW.
We must search in
(-radius + C) / 2..=(radius + C) / 2. However, this
makes the assumption that there are any matches. It is possible that the
radius is low enough that we get no matches. In this case we can test the
0 slope case. We just need to test if
TL = (radius + C) / 2 is
actually a match. To test that:
abs((radius + C) / 2 - SL) + abs(TW - (radius + C) / 2 - SR) <= radius.
If the test succeeds, then we can safely iterate over the range.
Lets apply this reasoning to the previously mentioned tree. We expect to
get the range
C = SL - SR + TW = 3 - 2 + 4 = 5
Now we need to test
abs((radius + C) / 2 - SL) + abs(TW - (radius + C) / 2 - SR) <= radius.
abs((1 + 5) / 2 - 3) + abs(4 - (1 + 5) / 2 - 2) <= 1
abs(6 / 2 - 3) + abs(4 - 6 / 2 - 2) <= 1
abs(0) + abs(-1) <= 1
1 <= 1
The test passes. Now we compute the range.
(-radius + C) / 2..=(radius + C) / 2
(-1 + 5) / 2..=(1 + 5) / 2
4 / 2..=6 / 2
This is the range we expected.
We may need to clip the range to be inside the bucket as well, since the radius might cover a bigger set of hamming distances than the range.
Now we wish to find all combinations of substrings that result in getting
below the radius. To do this we need to know the
SOD at each index we
search in a given substring. To do that we must describe the relationship
There are three phases in the iteration pattern over
TL. The first is
radius is going down, the second is when it stays flat, the
third is when it is going up. The test in the last part made sure the
bottom was above the radius. We need to compute the points at which the
slope becomes 0, which are the inflection points. Luckily, these are
trivial to calculate. They are when the inside of the
SOD is equal to
TL - SL = 0
TR - SR = 0
We also know that
TR = TW - TL, so we can rewrite this in terms of
TW - TL - SR = 0
We care about
TL when we hit the inflection point:
TL = SL
TL = -SR + TW
We dont care which inflection point we hit first, we just want to know
where it is. We can just take the
max of these two
expressions to get the beginning and ending of the flat part of the curve.
Now we want to solve for the
SOD. Just like last time, we start with
being lower that
TR being higher than
(SL - TL) + (TW - TL - SR) = SOD
(TL - SL) + (SR - TW + TL) = SOD
We can simplify these to make it a bit clearer:
C = SL - SR + TW
-2TL + C = SOD
2TL - C = SOD
It starts by going down with a slope of
-2 and ends going up with a slope
2 just like we expect.
We can use this expression to compute the
SOD for each part of iteration.
Now the iteration is split into three parts:
(-radius + C) / 2..SL(
SOD = -2TL + C)
SL..-SR + TW(
SOD = -2SL + C)
-SR + TW..=(radius + C) / 2(
SOD = 2TL - C)
At this point we can compute the
SOD over all of our input indices. Now
we iterate over all input indices specificed, compute their
SOD, and then
perform a search over subsequent substrings by passing them a
new_radius = radius - SOD. This guarantees that all paths in that
substring also dont exceed the total
SOD for all substrings in the level.
When we use the above radius searching algorithm, we search every feature that could be at a particular radius or lower. Unfortunately, when we are searching for nearest neighbors in a hamming weight tree, we must search at radius 0, then radius 1, and so on. If we use the above algorithm, since hamming space has incredibly thick boundaries (see the paper Thick Boundaries in Binary Space and Their Influence on Nearest-Neighbor Search), it can be possible that a great proportion of the entire hamming space is equidistant with the nearest neighbor. This means that our search algorithm will make us test all of those places in the space if they have tables in the tree.