在大集合中高效地找到低汉明距离的二进制字符串

问题:

给定一个大的(约1亿)无符号32位整数列表、一个无符号32位整数输入值和一个最大的 汉明距离，返回输入值指定汉明距离内的所有列表成员。

实际的数据结构保持列表是开放的，性能要求决定了内存解决方案，建立数据结构的成本是次要的，低成本查询数据结构是关键。

例如:

For a maximum Hamming Distance of 1 (values typically will be quite small)


And input:
00001000100000000000000001111101


The values:
01001000100000000000000001111101
00001000100000000010000001111101


should match because there is only 1 position in which the bits are different.


11001000100000000010000001111101


should not match because 3 bit positions are different.

到目前为止我的想法是:

对于汉明距离为0的退化情况，只需使用排序列表并对特定的输入值进行二进制搜索。

如果汉明距离只有1，我可以翻转原始输入中的每个位，然后重复上面的32次。

如何有效地(不扫描整个列表)发现汉明距离 > 1的列表成员。

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How about sorting the list and then doing a binary search in that sorted list on the different possible values within you Hamming Distance?

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You could pre-compute every possible variation of your original list within the specified hamming distance, and store it in a bloom filter. This gives you a fast "NO" but not necessarily a clear answer about "YES."

For YES, store a list of all the original values associated with each position in the bloom filter, and go through them one at a time. Optimize the size of your bloom filter for speed / memory trade-offs.

Not sure if it all works exactly, but seems like a good approach if you've got runtime RAM to burn and are willing to spend a very long time in pre-computation.

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最佳答案

Question: What do we know about the Hamming distance d(x,y)?

Answer:

  1. It is non-negative: d(x,y) ≥ 0
  2. It is only zero for identical inputs: d(x,y) = 0 ⇔ x = y
  3. It is symmetric: d(x,y) = d(y,x)
  4. It obeys the triangle inequality, d(x,z) ≤ d(x,y) + d(y,z)

Question: Why do we care?

Answer: Because it means that the Hamming distance is a metric for a metric space. There are algorithms for indexing metric spaces.

You can also look up algorithms for "spatial indexing" in general, armed with the knowledge that your space is not Euclidean but it is a metric space. Many books on this subject cover string indexing using a metric such as the Hamming distance.

Footnote: If you are comparing the Hamming distance of fixed width strings, you may be able to get a significant performance improvement by using assembly or processor intrinsics. For example, with GCC (manual) you do this:

static inline int distance(unsigned x, unsigned y)
{
return __builtin_popcount(x^y);
}

If you then inform GCC that you are compiling for a computer with SSE4a, then I believe that should reduce to just a couple opcodes.

Edit: According to a number of sources, this is sometimes/often slower than the usual mask/shift/add code. Benchmarking shows that on my system, a C version outperform's GCC's __builtin_popcount by about 160%.

Addendum: I was curious about the problem myself, so I profiled three implementations: linear search, BK tree, and VP tree. Note that VP and BK trees are very similar. The children of a node in a BK tree are "shells" of trees containing points that are each a fixed distance from the tree's center. A node in a VP tree has two children, one containing all the points within a sphere centered on the node's center and the other child containing all the points outside. So you can think of a VP node as a BK node with two very thick "shells" instead of many finer ones.

The results were captured on my 3.2 GHz PC, and the algorithms do not attempt to utilize multiple cores (which should be easy). I chose a database size of 100M pseudorandom integers. Results are the average of 1000 queries for distance 1..5, and 100 queries for 6..10 and the linear search.

  • Database: 100M pseudorandom integers
  • Number of tests: 1000 for distance 1..5, 100 for distance 6..10 and linear
  • Results: Average # of query hits (very approximate)
  • Speed: Number of queries per second
  • Coverage: Average percentage of database examined per query
-- BK Tree --   -- VP Tree --   -- Linear --
Dist    Results Speed   Cov     Speed   Cov     Speed   Cov
1          0.90 3800     0.048% 4200     0.048%
2         11     300     0.68%   330     0.65%
3        130      56     3.8%     63     3.4%
4        970      18    12%       22    10%
5       5700       8.5  26%       10    22%
6       2.6e4      5.2  42%        6.0  37%
7       1.1e5      3.7  60%        4.1  54%
8       3.5e5      3.0  74%        3.2  70%
9       1.0e6      2.6  85%        2.7  82%
10      2.5e6      2.3  91%        2.4  90%
any                                             2.2     100%

In your comment, you mentioned:

I think BK-trees could be improved by generating a bunch of BK-trees with different root nodes, and spreading them out.

I think this is exactly the reason why the VP tree performs (slightly) better than the BK tree. Being "deeper" rather than "shallower", it compares against more points rather than using finer-grained comparisons against fewer points. I suspect that the differences are more extreme in higher dimensional spaces.

A final tip: leaf nodes in the tree should just be flat arrays of integers for a linear scan. For small sets (maybe 1000 points or fewer) this will be faster and more memory efficient.

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One possible approach to solve this problem is using a Disjoint-set data structure. The idea is merge list members with Hamming distance <= k in the same set. Here is the outline of the algorithm:

  • For each list member calculate every possible value with Hamming distance <= k. For k=1, there are 32 values (for 32-bit values). For k=2, 32 + 32*31/2 values.

    • For each calculated value, test if it is in the original input. You can use an array with size 2^32 or a hash map to do this check.

    • If the value is in the original input, do a "union" operation with the list member.

    • Keep the number of union operations executed in a variable.

You start the algorithm with N disjoint sets (where N is the number of elements in the input). Each time you execute an union operation, you decrease by 1 the number of disjoint sets. When the algorithm terminates, the disjoint-set data structure will have all the values with Hamming distance <= k grouped in disjoint sets. This disjoint-set data structure can be calculated in almost linear time.

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I wrote a solution where I represent the input numbers in a bitset of 232 bits, so I can check in O(1) whether a certain number is in the input. Then for a queried number and maximum distance, I recursively generate all numbers within that distance and check them against the bitset.

For example for maximum distance 5, this is 242825 numbers (sumd = 0 to 5 {32 choose d}). For comparison, Dietrich Epp's VP-tree solution for example goes through 22% of the 100 million numbers, i.e., through 22 million numbers.

I used Dietrich's code/solutions as the basis to add my solution and compare it with his. Here are speeds, in queries per second, for maximum distances up to 10:

Dist     BK Tree     VP Tree         Bitset   Linear


1   10,133.83   15,773.69   1,905,202.76   4.73
2      677.78    1,006.95     218,624.08   4.70
3      113.14      173.15      27,022.32   4.76
4       34.06       54.13       4,239.28   4.75
5       15.21       23.81         932.18   4.79
6        8.96       13.23         236.09   4.78
7        6.52        8.37          69.18   4.77
8        5.11        6.15          23.76   4.68
9        4.39        4.83           9.01   4.47
10        3.69        3.94           2.82   4.13


Prepare     4.1s       21.0s          1.52s  0.13s
times (for building the data structure before the queries)

For small distances, the bitset solution is by far the fastest of the four. Question author Eric commented below that the largest distance of interest would probably be 4-5. Naturally, my bitset solution becomes slower for larger distances, even slower than the linear search (for distance 32, it would go through 232 numbers). But for distance 9 it still easily leads.

I also modified Dietrich's testing. Each of the above results is for letting the algorithm solve at least three queries and as many queries as it can in about 15 seconds (I do rounds with 1, 2, 4, 8, 16, etc queries, until at least 10 seconds have passed in total). That's fairly stable, I even get similar numbers for just 1 second.

My CPU is an i7-6700. My code (based on Dietrich's) is here (ignore the documentation there at least for now, not sure what to do about that, but the tree.c contains all the code and my test.bat shows how I compiled and ran (I used the flags from Dietrich's Makefile)). Shortcut to my solution.

One caveat: My query results contain numbers only once, so if the input list contains duplicate numbers, that may or may not be desired. In question author Eric's case, there were no duplicates (see comment below). In any case, this solution might be good for people who either have no duplicates in the input or don't want or need duplicates in the query results (I think it's likely that the pure query results are only a means to an end and then some other code turns the numbers into something else, for example a map mapping a number to a list of files whose hash is that number).

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A common approach (at least common to me) is to divide your bit string in several chunks and query on these chunks for an exact match as pre-filter step. If you work with files, you create as many files as you have chunks (e.g. 4 here) with each chunk permuted in front and then sort the files. You can use a binary search and you can even expand you search above and below a matching chunk for bonus.

You then can perform a bitwise hamming distance computation on the returned results which should be only a smaller subset of your overall dataset. This can be done using data files or SQL tables.

So to recap: Say you have a bunch of 32 bits strings in a DB or files and that you want to find every hash that are within a 3 bits hamming distance or less of your "query" bit string:

  1. create a table with four columns: each will contain an 8 bits (as a string or int) slice of the 32 bits hashes, islice 1 to 4. Or if you use files, create four files, each being a permutation of the slices having one "islice" at the front of each "row"

  2. slice your query bit string the same way in qslice 1 to 4.

  3. query this table such that any of qslice1=islice1 or qslice2=islice2 or qslice3=islice3 or qslice4=islice4. This gives you every string that are within 7 bits (8 - 1) of the query string. If using a file, do a binary search in each of the four permuted files for the same results.

  4. for each returned bit string, compute the exact hamming distance pair-wise with you query bit string (reconstructing the index-side bit strings from the four slices either from the DB or from a permuted file)

The number of operations in step 4 should be much less than a full pair-wise hamming computation of your whole table and is very efficient in practice. Furthermore, it is easy to shard the files in smaller files as need for more speed using parallelism.

Now of course in your case, you are looking for a self-join of sort, that is all the values that are within some distance of each other. The same approach still works IMHO, though you will have to expand up and down from a starting point for permutations (using files or lists) that share the starting chunk and compute the hamming distance for the resulting cluster.

If running in memory instead of files, your 100M 32 bits strings data set would be in the range of 4 GB. Hence the four permuted lists may need about 16GB+ of RAM. Though I get excellent results with memory mapped files instead and must less RAM for similar size datasets.

There are open source implementations available. The best in the space is IMHO the one done for Simhash by Moz, C++ but designed for 64 bits strings and not 32 bits.

This bounded happing distance approach was first described AFAIK by Moses Charikar in its "simhash" seminal paper and the corresponding Google patent:

  1. APPROXIMATE NEAREST NEIGHBOR SEARCH IN HAMMING SPACE

[...]

Given bit vectors consisting of d bits each, we choose N = O(n 1/(1+ ) ) random permutations of the bits. For each random permutation σ, we maintain a sorted order O σ of the bit vectors, in lexicographic order of the bits permuted by σ. Given a query bit vector q, we find the approximate nearest neighbor by doing the following:

For each permutation σ, we perform a binary search on O σ to locate the two bit vectors closest to q (in the lexicographic order obtained by bits permuted by σ). We now search in each of the sorted orders O σ examining elements above and below the position returned by the binary search in order of the length of the longest prefix that matches q.

Monika Henziger expanded on this in her paper "Finding near-duplicate web pages: a large-scale evaluation of algorithms":

3.3 The Results for Algorithm C

We partitioned the bit string of each page into 12 non- overlapping 4-byte pieces, creating 20B pieces, and computed the C-similarity of all pages that had at least one piece in common. This approach is guaranteed to find all pairs of pages with difference up to 11, i.e., C-similarity 373, but might miss some for larger differences.

This is also explained in the paper Detecting Near-Duplicates for Web Crawling by Gurmeet Singh Manku, Arvind Jain, and Anish Das Sarma:

  1. THE HAMMING DISTANCE PROBLEM

Definition: Given a collection of f -bit fingerprints and a query fingerprint F, identify whether an existing fingerprint differs from F in at most k bits. (In the batch-mode version of the above problem, we have a set of query fingerprints instead of a single query fingerprint)

[...]

Intuition: Consider a sorted table of 2 d f -bit truly random fingerprints. Focus on just the most significant d bits in the table. A listing of these d-bit numbers amounts to “almost a counter” in the sense that (a) quite a few 2 d bit- combinations exist, and (b) very few d-bit combinations are duplicated. On the other hand, the least significant f − d bits are “almost random”.

Now choose d such that |d − d| is a small integer. Since the table is sorted, a single probe suffices to identify all fingerprints which match F in d most significant bit-positions. Since |d − d| is small, the number of such matches is also expected to be small. For each matching fingerprint, we can easily figure out if it differs from F in at most k bit-positions or not (these differences would naturally be restricted to the f − d least-significant bit-positions).

The procedure described above helps us locate an existing fingerprint that differs from F in k bit-positions, all of which are restricted to be among the least significant f − d bits of F. This takes care of a fair number of cases. To cover all the cases, it suffices to build a small number of additional sorted tables, as formally outlined in the next Section.

Note: I posted a similar answer to a related DB-only question

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Here's a simple idea: do a byte-wise radix sort of the 100m input integers, most significant byte first, keeping track of bucket boundaries on the first three levels in some external structure.

To query, start with a distance budget of d and your input word w. For each bucket in the top level with byte value b, calculate the Hamming distance d_0 between b and the high byte of w. Recursively search that bucket with a budget of d - d_0: that is, for each byte value b', let d_1 be the Hamming distance between b' and the second byte of w. Recursively search into the third layer with a budget of w1, and so on.

Note that the buckets form a tree. Whenever your budget becomes negative, stop searching that subtree. If you recursively descend into a leaf without blowing your distance budget, that leaf value should be part of the output.

Here's one way to represent the external bucket boundary structure: have an array of length 16_777_216 (= (2**8)**3 = 2**24), where the element at index i is the starting index of the bucket containing values in range [256*i, 256*i + 255]. To find the index one beyond the end of that bucket, look up at index i+1 (or use the end of the array for i + 1 = 2**24).

Memory budget is 100m * 4 bytes per word = 400 MB for the inputs, and 2**24 * 4 bytes per address = 64 MiB for the indexing structure, or just shy of half a gig in total. The indexing structure is a 6.25% overhead on the raw data. Of course, once you've constructed the indexing structure you only need to store the lowest byte of each input word, since the other three are implicit in the index into the indexing structure, for a total of ~(64 + 50) MB.

If your input is not uniformly distributed, you could permute the bits of your input words with a (single, universally shared) permutation which puts all the entropy towards the top of the tree. That way, the first level of pruning will eliminate larger chunks of the search space.

I tried some experiments, and this performs about as well as linear search, sometimes even worse. So much for this fancy idea. Oh well, at least it's memory efficient.


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