As noted, this is an approximate algorithm. It gives an approximation for the number of unique objects, along with a standard deviation $\sigma$, which can then be used to determine bounds on the approximation with a desired maximum error $\epsilon$, if needed.

Given below are the following:

- intuition behind the algorithm
- the algorithm itself
- a java-based implementation
- some results using that implementation and
- some closing thoughts.

Intuition

If we had a good, random hash function that acted on strings and generated integers, what can we say about the generated integers? Since they are random themselves, we would expect:

- $1/2$ of them to have their binary representation end in $0$ (i.e. divisible by $2$),
- $1/4$ of them to have their binary representation end in $00$ (i.e. divisible by $4$)
- $1/8$ of them to have their binary representation end in $000$ (i.e. divisible by $8$)
- and in general, $1/2^n$ of them to have their binary representation end in $0^n$.

Turning the problem around, if the hash function generated an integer ending in $0^m$ bits (

To facilitate the above, this algorithm maintains 1 bit for each $0^i$ seen - i.e. 1 bit for 0, another for 00, another for 000, and so on. The output of the algorithm is based on the maximum of consecutive $0^i$ seen.

*and*it also generated integers ending in $0^{m-1}$ bits, $0^{m-2}$ bits, ..., $0^1$ bits), intuitively, the number of unique strings is around $2^m$.To facilitate the above, this algorithm maintains 1 bit for each $0^i$ seen - i.e. 1 bit for 0, another for 00, another for 000, and so on. The output of the algorithm is based on the maximum of consecutive $0^i$ seen.

The Flajolet-Martin algorithm

This is an informal description. Formal treatment can be found in the original paper listed in the reference section.

- Create a bit vector (bit array) of sufficient length $L$, such that $2^L>n$, the number of elements in the stream. Usually a 64-bit vector is sufficient since $2^{64}$ is quite large for most purposes.
- The i-th bit in this vector/array represents whether we have seen a hash function value whose binary representation ends in $0^i$. So initialize each bit to 0.
- Generate a good, random hash function that maps input (usually strings) to natural numbers.
- Read input. For each word, hash it and determine the number of trailing zeros. If the number of trailing zeros is k, set the k-th bit in the bit vector to 1.
- Once input is exhausted, get the index of the first 0 in the bit array (call this R). By the way, this is just the number of consecutive 1s (i.e. we have seen 0, 00, ..., as the output of the hash function) plus one.
- Calculate the number of unique words as $2^R/\phi$, where $\phi$ is 0.77351. A proof for this can be found in the original paper listed in the reference section.
- The standard deviation of R is a constant: $\sigma(R)=1.12$. (In other words, R can be off by about 1 for 1-0.68=32% of the observations, off by 2 for about 1-0.95=5% of the observations, off by 3 for 1-0.997=0.3% of the observations using the Empirical rule of statistics). This implies that our count can be off by a factor of 2 for 32% of the observations, off by a factory of 4 for 5% of the observations, off by a factor of 8 for 0.3% of the observations and so on.

- (Averaging) Use multiple hash functions and use the average R instead.
- (Bucketing) Averages are susceptible to large fluctuations. So use multiple buckets of hash functions from the above step and use the median of the average R. This gives fairly good accuracy.
- Overall accuracy of this algorithm can be tuned by using appropriate number of hash functions in the averaging and bucketing steps. Of course, if more accuracy is desired, more hash functions need to be used, which implies higher computation cost.

Java-based implementation

The code can be found here: FlajoletMartin.java. It uses lucene-core-3.0.3.jar or better.

Results using the above implementation

- Wikipedia article on "United States Constitution" had 3978 unique words. When run ten times, Flajolet-Martin algorithm reported values of 4902, 4202, 4202, 4044, 4367, 3602, 4367, 4202, 4202 and 3891 for an average of 4198. As can be seen, the average is about right, but the deviation is between -400 to 1000.
- Wikipedia article on "George Washington" had 3252 unique words. When run ten times, the reported values were 4044, 3466, 3466, 3466, 3744, 3209, 3335, 3209, 3891 and 3088, for an average of 3492.

Closing thoughts

- The Flajolet-Martin algorithm approximates the number of unique elements quite well, using just O(log m) memory, where m is the number of unique words.
- During implementation, it was observed that this algorithm is quite sensitive to the hash function parameters. The hash functions suggested in the original paper ($h(x)=(M+N\sum{ord(x_j)*128^j)\mod(2^L)}$) work properly only when n is odd. Otherwise, the algorithm always reports the number of unique elements as 1 or 2! I chose both $M$ and $N$ as odd, as can be seen in the implementation.

References

- Flajolet, P. and Martin, N.,
*Journal of Computer and System Sciences*, 1985. PDF - http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.81.3869&rep=rep1&type=pdf - J. Ullman and A. Rajaraman,
*Mining of Massive Datasets*,*Chapter 3*- available at http://infolab.stanford.edu/~ullman/mmds/ch4.pdf

great article :)

ReplyDeletethanks for the simple explanation!

ReplyDeleteA wonderful explanation. Only one thought : If you can't explain it simply, you don't understand it well enough. – Albert Einstein

ReplyDeleteYou have explained it very lucidly.

Thanks!!

Thanks for the article!

ReplyDeleteBeautiful Article. The intuition behind the algorithm is nicely illustrated.

ReplyDeleteGreat explanation! I'm a little confused about step 7: did you mean R can be off by one for

ReplyDelete1 -.68 = 32%, off by two for1 -.95 = 5%, and off by three for1 -.997 = 0.3% of observations?Yes!

DeleteThanks for your correction, Arel. I will update this article.

Great explanation.. I have a question though while combining estimates instead of taking average of 2 to the power R can we first take avergae of R then take 2 to power avg(R) will it solve the problem of taking large estimates

ReplyDeleteUltimate explanations thanks a lot..i was bit confuse but now all clear.

ReplyDelete