I was going through Eric Lippert's latest Blog post for Guidelines and rules for GetHashCode when i hit this para:
We could be even more clever here; just as a List resizes itself when it gets full, the bucket set could resize itself as well, to ensure that the average bucket length stays low. Also, for technical reasons it is often a good idea to make the bucket set length a prime number, rather than 100. There are plenty of improvements we could make to this hash table. But this quick sketch of a naive implementation of a hash table will do for now. I want to keep it simple.
So looks like i'm missing something. Why is it a good practice to set it to a prime number?.
You can find people that suggest the two opposite ends of the spectrum. On the one side, choosing a prime number for the size of the hash table will reduce the chances of collisions, even if the hash function is not too effective distributing the results. Note that if (in the simplest example to argue about) a power of 2 size is decided, only the lower bits affect the bucket, while for a prime number most bits in the result of the hash will be used.
On the other hand, you can gain more by choosing a better hash function, or even rehashing he result of the hash function by applying some bit operations, and using a power of 2 hash size to speed up calculations.
As an example from real life, Java HashTable were initially implemented by using prime (or almost prime sizes), but from Java 1.4 on, the design was changed to use power of two number of buckets and added a second fast hash function applied to the result of the initial hash. An interesting article commenting that change can be found here.
So basically:
a prime number helps dispersing the inputs across the different buckets even in the event of not-so-good hash functions.
a similar effect can be achieved by post processing the result of the hash function, and using a power of 2 size to speedup the modulo operation (bit mask) and compensate for the post processing.
Because this produces a better hash function and reduces the number of possible collisions. This is explained in Choosing a good hashing function:
A basic requirement is that the
function should provide a uniform
distribution of hash values. A
non-uniform distribution increases the
number of collisions, and the cost of
resolving them.
The distribution needs to be uniform
only for table sizes s that occur in
the application. In particular, if one
uses dynamic resizing with exact
doubling and halving of s, the hash
function needs to be uniform only when
s is a power of two. On the other
hand, some hashing algorithms provide
uniform hashes only when s is a prime
number.
Say your bucket set length is a power of 2 - that makes the mod calculations quite fast. It also means that the bucket selection is determine solely by the top m bits of the hash code. (Where m = 32 - n, where n is the power of 2 being used). So it's like you're throwing away useful bits of the hashcode immediately.
Or as in this blog post from 2006 puts it:
Suppose your hashCode function results in the following hashCodes among others {x , 2x, 3x, 4x, 5x, 6x...}, then all these are going to be clustered in just m number of buckets, where m = table_length/GreatestCommonFactor(table_length, x). (It is trivial to verify/derive this). Now you can do one of the following to avoid clustering:
...
Or simply make m equal to the table_length by making GreatestCommonFactor(table_length, x) equal to 1, i.e by making table_length coprime with x. And if x can be just about any number then make sure that table_length is a prime number.
Related
I am bullding a hash table using double hashing to solve collision. How can I know what is the propper size? I know it has to prime to minmize the number of collisions.
The easiest way to implement hash tables is to use a power-of-2 sized hash tables.
The reason is that if N = 2M, then calculating H % N is as simple as calculating H & (N - 1).
With fast hash functions such as MurmurHash3_32, the slowest part of using the hash table is actually calculating the modulo. H & (N - 1) does not calculate modulo but bitwise AND which is much faster (and it's the same as modulo if N is a power of 2).
Somebody could validly claim that MurmurHash suffers from seed-independent multicollisions and therefore is susceptible to a hash collision denial of service attack. That's true, but you shouldn't use linked lists to resolve hash collisions. You should use only hash tables where the keys are sortable by some comparison function (larger than, equal, smaller than) and then you can use red-black trees (or AVL trees) to resolve hash collisions. If there's no natural comparison functions (such as for complex numbers), you can invent one.
Using a red-black tree that almost always is just a single root element with MurmurHash is much faster than trying to be "secure" by using SipHash and then stupidly using linked lists to resolve hash collisions (which caused the need for the absurdly slow SipHash in the first place).
In theory, with non-power-of-2-sized hash tables where the size is rarely varying, you could use the "fast division by invariant integers using multiplication" trick but that's slower than power-of-2-sizing and bitwise AND.
The prime sizing is just for really poor hash functions. MurmurHash, although it suffers from seed-independent multicollisions, does not suffer from collisions with reasonable (non-attacker-generated) keys if the table size is a power of 2.
No, there is no point in making the size be prime and that adds a lot of extra work for you. Just make the size be a power of two and double it whenever the number of objects in the hash table reaches some threshold, like 50% or 25% of the size.
If you are asking about the current size, you may use the sizeof(table)/sizeof(element) function since you are using the double hashing method.
If you are asking about the new size of the hash table once full (passed a certain criterion), then the most common is to add 10 new slots. This should be based on what are you using your table for. The default setting on most built in tables in other languages is if 0.75 full, then add 10 slots.
If it's about something else, then please modify your question, so it's more descriptive.
Edit: I just noticed the answer above me and I think that using the 2^p method is very common too in exponentially increasing tables and is very helpful with double hashing.
I need a hash function that is as efficient as possible, for a hash table (actually a hash set) that uses probing (open addressing) for collision resolution. The entries stored in the table are all 4 byte ints that take on random values over the range.
I am considering something even faster than djb2, something like
value mod LARGE_PRIME
Then mod it again with my bucket size. I suppose this prime is necessarily bigger than my bucket size, which means I also have some kind of sanity limit on how big my table has to grow (it probably won't ever get past 256 entries).
I don't require any cryptography aspects of hash function - as long as it isn't terribly collision-prone, it should work fine.
Will this make a good hash function? Can I define a specific algorithm for my hash table capacity every time I resize to improve it?
The right hash function boils down to the data you are hashing: how random are your values? If your values are uniformly distributed over the range, and the range is much larger than the number of hash buckets, then just using
value MOD number_of_buckets
will be a reasonable hash function - adding in MOD <prime> won't actually give you very much, and in fact might well make the hashing worse (because some buckets will be under- or over-used more than they would otherwise have been).
Primes aren't magic - they can sometimes be used to "smooth out" correlation effects due to common factors, but if you don't have those correlations to begin with, you may be better off without them - especially if speed is paramount!
I've been learning about different algorithms in my spare time recently, and one that I came across which appears to be very interesting is called the HyperLogLog algorithm - which estimates how many unique items are in a list.
This was particularly interesting to me because it brought me back to my MySQL days when I saw that "Cardinality" value (which I always assumed until recently that it was calculated not estimated).
So I know how to write an algorithm in O(n) that will calculate how many unique items are in an array. I wrote this in JavaScript:
function countUniqueAlgo1(arr) {
var Table = {};
var numUnique = 0;
var numDataPoints = arr.length;
for (var j = 0; j < numDataPoints; j++) {
var val = arr[j];
if (Table[val] != null) {
continue;
}
Table[val] = 1;
numUnique++;
}
return numUnique;
}
But the problem is that my algorithm, while O(n), uses a lot of memory (storing values in Table).
I've been reading this paper about how to count duplicates in a list in O(n) time and using minimal memory.
It explains that by hashing and counting bits or something one can estimate within a certain probability (assuming the list is evenly distributed) the number of unique items in a list.
I've read the paper, but I can't seem to understand it. Can someone give a more layperson's explanation? I know what hashes are, but I don't understand how they are used in this HyperLogLog algorithm.
The main trick behind this algorithm is that if you, observing a stream of random integers, see an integer which binary representation starts with some known prefix, there is a higher chance that the cardinality of the stream is 2^(size of the prefix).
That is, in a random stream of integers, ~50% of the numbers (in binary) starts with "1", 25% starts with "01", 12,5% starts with "001". This means that if you observe a random stream and see a "001", there is a higher chance that this stream has a cardinality of 8.
(The prefix "00..1" has no special meaning. It's there just because it's easy to find the most significant bit in a binary number in most processors)
Of course, if you observe just one integer, the chance this value is wrong is high. That's why the algorithm divides the stream in "m" independent substreams and keep the maximum length of a seen "00...1" prefix of each substream. Then, estimates the final value by taking the mean value of each substream.
That's the main idea of this algorithm. There are some missing details (the correction for low estimate values, for example), but it's all well written in the paper. Sorry for the terrible english.
A HyperLogLog is a probabilistic data structure. It counts the number of distinct elements in a list. But in comparison to a straightforward way of doing it (having a set and adding elements to the set) it does this in an approximate way.
Before looking how the HyperLogLog algorithm does this, one has to understand why you need it. The problem with a straightforward way is that it consumes O(distinct elements) of space. Why there is a big O notation here instead of just distinct elements? This is because elements can be of different sizes. One element can be 1 another element "is this big string". So if you have a huge list (or a huge stream of elements) it will take a lot memory.
Probabilistic Counting
How can one get a reasonable estimate of a number of unique elements? Assume that you have a string of length m which consists of {0, 1} with equal probability. What is the probability that it will start with 0, with 2 zeros, with k zeros? It is 1/2, 1/4 and 1/2^k. This means that if you have encountered a string starting with k zeros, you have approximately looked through 2^k elements. So this is a good starting point. Having a list of elements that are evenly distributed between 0 and 2^k - 1 you can count the maximum number of the biggest prefix of zeros in binary representation and this will give you a reasonable estimate.
The problem is that the assumption of having evenly distributed numbers from 0 t 2^k-1 is too hard to achieve (the data we encountered is mostly not numbers, almost never evenly distributed, and can be between any values. But using a good hashing function you can assume that the output bits would be evenly distributed and most hashing function have outputs between 0 and 2^k - 1 (SHA1 give you values between 0 and 2^160). So what we have achieved so far is that we can estimate the number of unique elements with the maximum cardinality of k bits by storing only one number of size log(k) bits. The downside is that we have a huge variance in our estimate. A cool thing that we almost created 1984's probabilistic counting paper (it is a little bit smarter with the estimate, but still we are close).
LogLog
Before moving further, we have to understand why our first estimate is not that great. The reason behind it is that one random occurrence of high frequency 0-prefix element can spoil everything. One way to improve it is to use many hash functions, count max for each of the hash functions and in the end average them out. This is an excellent idea, which will improve the estimate, but LogLog paper used a slightly different approach (probably because hashing is kind of expensive).
They used one hash but divided it into two parts. One is called a bucket (total number of buckets is 2^x) and another - is basically the same as our hash. It was hard for me to get what was going on, so I will give an example. Assume you have two elements and your hash function which gives values form 0 to 2^10 produced 2 values: 344 and 387. You decided to have 16 buckets. So you have:
0101 011000 bucket 5 will store 1
0110 000011 bucket 6 will store 4
By having more buckets you decrease the variance (you use slightly more space, but it is still tiny). Using math skills they were able to quantify the error (which is 1.3/sqrt(number of buckets)).
HyperLogLog
HyperLogLog does not introduce any new ideas, but mostly uses a lot of math to improve the previous estimate. Researchers have found that if you remove 30% of the biggest numbers from the buckets you significantly improve the estimate. They also used another algorithm for averaging numbers. The paper is math-heavy.
And I want to finish with a recent paper, which shows an improved version of hyperLogLog algorithm (up until now I didn't have time to fully understand it, but maybe later I will improve this answer).
The intuition is if your input is a large set of random number (e.g. hashed values), they should distribute evenly over a range. Let's say the range is up to 10 bit to represent value up to 1024. Then observed the minimum value. Let's say it is 10. Then the cardinality will estimated to be about 100 (10 × 100 ≈ 1024).
Read the paper for the real logic of course.
Another good explanation with sample code can be found here:
Damn Cool Algorithms: Cardinality Estimation - Nick's Blog
Say, i have 10 billions of numbers stored in a file. How would i find the number that has already appeared once previously?
Well i can't just populate billions of number at a stretch in array and then keep a simple nested loop to check if the number has appeared previously.
How would you approach this problem?
Thanks in advance :)
I had this as an interview question once.
Here is an algorithm that is O(N)
Use a hash table. Sequentially store pointers to the numbers, where the hash key is computed from the number value. Once you have a collision, you have found your duplicate.
Author Edit:
Below, #Phimuemue makes the excellent point that 4-byte integers have a fixed bound before a collision is guaranteed; that is 2^32, or approx. 4 GB. When considered in the conversation accompanying this answer, worst-case memory consumption by this algorithm is dramatically reduced.
Furthermore, using the bit array as described below can reduce memory consumption to 1/8th, 512mb. On many machines, this computation is now possible without considering either a persistent hash, or the less-performant sort-first strategy.
Now, longer numbers or double-precision numbers are less-effective scenarios for the bit array strategy.
Phimuemue Edit:
Of course one needs to take a bit "special" hash table:
Take a hashtable consisting of 2^32 bits. Since the question asks about 4-byte-integers, there are at most 2^32 different of them, i.e. one bit for each number. 2^32 bit = 512mb.
So now one has just to determine the location of the corresponding bit in the hashmap and set it. If one encounters a bit which already is set, the number occured in the sequence already.
The important question is whether you want to solve this problem efficiently, or whether you want accurately.
If you truly have 10 billion numbers and just one single duplicate, then you are in a "needle in the haystack" type of situation. Intuitively, short of very grimy and unstable solution, there is no hope of solving this without storing a significant amount of the numbers.
Instead, turn to probabilistic solutions, which have been used in most any practical application of this problem (in network analysis, what you are trying to do is look for mice, i.e., elements which appear very infrequently in a large data set).
A possible solution, which can be made to find exact results: use a sufficiently high-resolution Bloom filter. Either use the filter to determine if an element has already been seen, or, if you want perfect accuracy, use (as kbrimington suggested you use a standard hash table) the filter to, eh, filter out elements which you can't possibly have seen and, on a second pass, determine the elements you actually see twice.
And if your problem is slightly different---for instance, you know that you have at least 0.001% elements which repeat themselves twice, and you would like to find out how many there are approximately, or you would like to get a random sample of such elements---then a whole score of probabilistic streaming algorithms, in the vein of Flajolet & Martin, Alon et al., exist and are very interesting (not to mention highly efficient).
Read the file once, create a hashtable storing the number of times you encounter each item. But wait! Instead of using the item itself as a key, you use a hash of the item iself, for example the least significant digits, let's say 20 digits (1M items).
After the first pass, all items that have counter > 1 may point to a duplicated item, or be a false positive. Rescan the file, consider only items that may lead to a duplicate (looking up each item in table one), build a new hashtable using real values as keys now and storing the count again.
After the second pass, items with count > 1 in the second table are your duplicates.
This is still O(n), just twice as slow as a single pass.
How about:
Sort input by using some algorith which allows only portion of input to be in RAM. Examples are there
Seek duplicates in output of 1st step -- you'll need space for just 2 elements of input in RAM at a time to detect repetitions.
Finding duplicates
Noting that its a 32bit integer means that you're going to have a large number of duplicates, since a 32 bit int can only represent 4.3ish billion different numbers and you have "10 billions".
If you were to use a tightly packed set you could represent whether all the possibilities are in 512 MB, which can easily fit into current RAM values. This as a start pretty easily allows you to recognise the fact if a number is duplicated or not.
Counting Duplicates
If you need to know how many times a number is duplicated you're getting into having a hashmap that contains only duplicates (using the first 500MB of the ram to tell efficiently IF it should be in the map or not). At a worst case scenario with a large spread you're not going to be able fit that into ram.
Another approach if the numbers will have an even amount of duplicates is to use a tightly packed array with 2-8 bits per value, taking about 1-4GB of RAM allowing you to count up to 255 occurrances of each number.
Its going to be a hack, but its doable.
You need to implement some sort of looping construct to read the numbers one at a time since you can't have them in memory all at once.
How? Oh, what language are you using?
You have to read each number and store it into a hashmap, so that if a number occurs again, it will automatically get discarded.
If possible range of numbers in file is not too large then you can use some bit array to indicate if some of the number in range appeared.
If the range of the numbers is small enough, you can use a bit field to store if it is in there - initialize that with a single scan through the file. Takes one bit per possible number.
With large range (like int) you need to read through the file every time. File layout may allow for more efficient lookups (i.e. binary search in case of sorted array).
If time is not an issue and RAM is, you could read each number and then compare it to each subsequent number by reading from the file without storing it in RAM. It will take an incredible amount of time but you will not run out of memory.
I have to agree with kbrimington and his idea of a hash table, but first of all, I would like to know the range of the numbers that you're looking for. Basically, if you're looking for 32-bit numbers, you would need a single array of 4.294.967.296 bits. You start by setting all bits to 0 and every number in the file will set a specific bit. If the bit is already set then you've found a number that has occurred before. Do you also need to know how often they occur?Still, it would need 536.870.912 bytes at least. (512 MB.) It's a lot and would require some crafty programming skills. Depending on your programming language and personal experience, there would be hundreds of solutions to solve it this way.
Had to do this a long time ago.
What i did... i sorted the numbers as much as i could (had a time-constraint limit) and arranged them like this while sorting:
1 to 10, 12, 16, 20 to 50, 52 would become..
[1,10], 12, 16, [20,50], 52, ...
Since in my case i had hundreds of numbers that were very "close" ($a-$b=1), from a few million sets i had a very low memory useage
p.s. another way to store them
1, -9, 12, 16, 20, -30, 52,
when i had no numbers lower than zero
After that i applied various algorithms (described by other posters) here on the reduced data set
#include <stdio.h>
#include <stdlib.h>
/* Macro is overly general but I left it 'cos it's convenient */
#define BITOP(a,b,op) \
((a)[(size_t)(b)/(8*sizeof *(a))] op (size_t)1<<((size_t)(b)%(8*sizeof *(a))))
int main(void)
{
unsigned x=0;
size_t *seen = malloc(1<<8*sizeof(unsigned)-3);
while (scanf("%u", &x)>0 && !BITOP(seen,x,&)) BITOP(seen,x,|=);
if (BITOP(seen,x,&)) printf("duplicate is %u\n", x);
else printf("no duplicate\n");
return 0;
}
This is a simple problem that can be solved very easily (several lines of code) and very fast (several minutes of execution) with the right tools
my personal approach would be in using MapReduce
MapReduce: Simplified Data Processing on Large Clusters
i'm sorry for not going into more details but once getting familiar with the concept of MapReduce it is going to be very clear on how to target the solution
basicly we are going to implement two simple functions
Map(key, value)
Reduce(key, values[])
so all in all:
open file and iterate through the data
for each number -> Map(number, line_index)
in the reduce we will get the number as the key and the total occurrences as the number of values (including their positions in the file)
so in Reduce(key, values[]) if number of values > 1 than its a duplicate number
print the duplicates : number, line_index1, line_index2,...
again this approach can result in a very fast execution depending on how your MapReduce framework is set, highly scalable and very reliable, there are many diffrent implementations for MapReduce in many languages
there are several top companies presenting already built up cloud computing environments like Google, Microsoft azure, Amazon AWS, ...
or you can build your own and set a cluster with any providers offering virtual computing environments paying very low costs by the hour
good luck :)
Another more simple approach could be in using bloom filters
AdamT
Implement a BitArray such that ith index of this array will correspond to the numbers 8*i +1 to 8*(i+1) -1. ie first bit of ith number is 1 if we already had seen 8*i+1. Second bit of ith number is 1 if we already have seen 8*i + 2 and so on.
Initialize this bit array with size Integer.Max/8 and whenever you saw a number k, Set the k%8 bit of k/8 index as 1 if this bit is already 1 means you have seen this number already.
I have some data, up to a between a million and a billion records, each which is represented by a bitfield, about 64 bits per key. The bits are independent, you can imagine them basically as random bits.
If I have a test key and I want to find all values in my data with the same key, a hash table will spit those out very easily, in O(1).
What algorithm/data structure would efficiently find all records most similar to the query key? Here similar means that most bits are identical, but a minimal number are allowed to be wrong. This is traditionally measured by Hamming distance., which just counts the number of mismatched bits.
There's two ways this query might be made, one might be by specifying a mismatch rate like "give me a list of all existing keys which have less than 6 bits that differ from my query" or by simply best matches, like "give me a list of the 10,000 keys which have the lowest number of differing bits from my query."
You might be temped to run to k-nearest-neighbor algorithms, but here we're talking about independent bits, so it doesn't seem likely that structures like quadtrees are useful.
The problem can be solved by simple brute force testing a hash table for low numbers of differing bits. If we want to find all keys that differ by one bit from our query, for example, we can enumerate all 64 possible keys and test them all. But this explodes quickly, if we wanted to allow two bits of difference, then we'd have to probe 64*63=4032 times. It gets exponentially worse for higher numbers of bits.
So is there another data structure or strategy that makes this kind of query more efficient?
The database/structure can be preprocessed as much as you like, it's the query speed that should be optimized.
What you want is a BK-Tree. It's a tree that's ideally suited to indexing metric spaces (your problem is one), and supports both nearest-neighbour and distance queries. I wrote an article about it a while ago.
BK-Trees are generally described with reference to text and using levenshtein distance to build the tree, but it's straightforward to write one in terms of binary strings and hamming distance.
This sounds like a good fit for an S-Tree, which is like a hierarchical inverted file. Good resources on this topic include the following papers:
Hierarchical Bitmap Index: An Efficient and Scalable Indexing Technique for Set-Valued Attributes.
Improved Methods for Signature-Tree Construction (2000)
Quote from the first one:
The hierarchical bitmap index efficiently supports dif-
ferent classes of queries, including subset, superset and similarity queries.
Our experiments show that the hierarchical bitmap index outperforms
other set indexing techniques significantly.
These papers include references to other research that you might find useful, such as M-Trees.
Create a binary tree (specifically a trie) representing each key in your start set in the following way: The root node is the empty word, moving down the tree to the left appends a 0 and moving down the right appends a 1. The tree will only have as many leaves as your start set has elements, so the size should stay manageable.
Now you can do a recursive traversal of this tree, allowing at most n "deviations" from the query key in each recursive line of execution, until you have found all of the nodes in the start set which are within that number of deviations.
I'd go with an inverted index, like a search engine. You've basically got a fixed vocabulary of 64 words. Then similarity is measured by hamming distance, instead of cosine similarity like a search engine would want to use. Constructing the index will be slow, but you ought to be able to query it with normal search enginey speeds.
The book Introduction to Information Retrieval covers the efficient construction, storage, compression and querying of inverted indexes.
"Near-optimal hashing algorithms for approximate nearest neighbor in high dimensions", from 2008, seems to be the best result as of then. I won't try to summarize since I read it over a year ago and it's hairy. That's from a page on locality-sensitive hashing, along with an implementation of an earlier version of the scheme. For more general pointers, read up on nearest neighbor search.
This kind of question has been asked before: Fastest way to find most similar string to an input?
The database/structure can be
preprocessed as much as you like
Well...IF that is true. Then all you need is a similarity matrix of your hamming distances. Make the matrix sparse by pruning out large distances. It doesn't get any faster and not that much of a memory hog.
Well, you could insert all of the neighbor keys along with the original key. That would mean that you store (64 choose k) times as much data, for k differing bits, and it will require that you decide k beforehand. Though you could always extend k by brute force querying neighbors, and this will automatically query the neighbors of your neighbors that you inserted. This also gives you a time-space tradeoff: for example, if you accept a 64 x data blowup and 64 times slower you can get two bits of distance.
I haven't completely thought this through, but I have an idea of where I'd start.
You could divide the search space up into a number of buckets where each bucket has a bucket key and the keys in the bucket are the keys that are more similar to this bucket key than any other bucket key. To create the bucket keys, you could randomly generate 64 bit keys and discard any that are too close to any previously created bucket key, or you could work out some algorithm that generates keys that are all dissimilar enough. To find the closest key to a test key, first find the bucket key that is closest, and then test each key in the bucket. (Actually, it's possible, but not likely, for the closest key to be in another bucket - do you need to find the closest key, or would a very close key be good enough?)
If you're ok with doing it probabilistically, I think there's a good way to solve question 2. I assume you have 2^30 data and cutoff and you want to find all points within cutoff distance from test.
One_Try()
1. Generate randomly a 20-bit subset S of 64 bits
2. Ask for a list of elements that agree with test on S (about 2^10 elements)
3. Sort that list by Hamming distance from test
4. Discard the part of list after cutoff
You repeat One_Try as much as you need while merging the lists. The more tries you have, the more points you find. For example, if x is within 5 bits, you'll find it in one try with about (2/3)^5 = 13% probability. Therefore if you repeat 100 tries you find all but roughly 10^{-6} of such x. Total time: 100*(1000*log 1000).
The main advantage of this is that you're able to output answers to question 2 as you proceed, since after the first few tries you'll certainly find everything within distance not more than 3 bits, etc.
If you have many computers, you give each of them several tries, since they are perfectly parallelizable: each computer saves some hash tables in advance.
Data structures for large sets described here: Detecting Near-Duplicates for Web Crawling
or
in memory trie: Judy-arrays at sourceforge.net
Assuming you have to visit each row to test its value (or if you index on the bitfield then each index entry), then you can write the actual test quite efficiently using
A xor B
To find the difference bits, then bit-count the result, using a technique like this.
This effectively gives you the hamming distance.
Since this can compile down to tens of instructions per test, this can run pretty fast.
If you are okay with a randomized algorithm (monte carlo in this case), you can use the minhash.
If the data weren't so sparse, a graph with keys as the vertices and edges linking 'adjacent' (Hamming distance = 1) nodes would probably be very efficient time-wise. The space would be very large though, so in your case, I don't think it would be a worthwhile tradeoff.