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algorithm - How many hash functions does my bloom filter need?

Wikipedia says:

An empty Bloom filter is a bit array of m bits, all set to 0. There must also be k different hash functions defined, each of which maps or hashes some set element to one of the m array positions with a uniform random distribution.

I read the article, but what I don't understand is how k is determined. Is it a function of the table size?

Also, in hash tables I've written I used a simple but effective algorithm for automatically growing the hash's size. Basically, if ever more than 50% of the buckets in the table were filled, I would double the size of the table. I suspect you might still want to do this with a bloom filter to reduce false positives. Correct ?

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Given:

  • n: how many items you expect to have in your filter (e.g. 216,553)
  • p: your acceptable false positive rate {0..1} (e.g. 0.01 → 1%)

we want to calculate:

  • m: the number of bits needed in the bloom filter
  • k: the number of hash functions we should apply

The formulas:

m = -n*ln(p) / (ln(2)^2) the number of bits
k = m/n * ln(2) the number of hash functions

In our case:

  • m = -216553*ln(0.01) / (ln(2)^2) = 997263 / 0.48045 = 2,075,686 bits (253 kB)
  • k = m/n * ln(2) = 2075686/216553 * 0.693147 = 6.46 hash functions (7 hash functions)

Note: Any code released into public domain. No attribution required.


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