The Algorithms logo
The Algorithms

Bloom Filter


The use of this data structure is to test membership in a set.
Compared to Python's built-in set() it is more space-efficient.
In the following example, only 8 bits of memory will be used:
>>> bloom = Bloom(size=8)

Initially, the filter contains all zeros:
>>> bloom.bitstring

When an element is added, two bits are set to 1
since there are 2 hash functions in this implementation:
>>> "Titanic" in bloom
>>> bloom.add("Titanic")
>>> bloom.bitstring
>>> "Titanic" in bloom

However, sometimes only one bit is added
because both hash functions return the same value
>>> bloom.add("Avatar")
>>> "Avatar" in bloom
>>> bloom.format_hash("Avatar")
>>> bloom.bitstring

Not added elements should return False ...
>>> not_present_films = ("The Godfather", "Interstellar", "Parasite", "Pulp Fiction")
>>> {
...   film: bloom.format_hash(film) for film in not_present_films
... } # doctest: +NORMALIZE_WHITESPACE
{'The Godfather': '00000101',
 'Interstellar': '00000011',
 'Parasite': '00010010',
 'Pulp Fiction': '10000100'}
>>> any(film in bloom for film in not_present_films)

but sometimes there are false positives:
>>> "Ratatouille" in bloom
>>> bloom.format_hash("Ratatouille")

The probability increases with the number of elements added.
The probability decreases with the number of bits in the bitarray.
>>> bloom.estimated_error_rate
>>> bloom.add("The Godfather")
>>> bloom.estimated_error_rate
>>> bloom.bitstring

from hashlib import md5, sha256

HASH_FUNCTIONS = (sha256, md5)

class Bloom:
    def __init__(self, size: int = 8) -> None:
        self.bitarray = 0b0
        self.size = size

    def add(self, value: str) -> None:
        h = self.hash_(value)
        self.bitarray |= h

    def exists(self, value: str) -> bool:
        h = self.hash_(value)
        return (h & self.bitarray) == h

    def __contains__(self, other: str) -> bool:
        return self.exists(other)

    def format_bin(self, bitarray: int) -> str:
        res = bin(bitarray)[2:]
        return res.zfill(self.size)

    def bitstring(self) -> str:
        return self.format_bin(self.bitarray)

    def hash_(self, value: str) -> int:
        res = 0b0
        for func in HASH_FUNCTIONS:
            position = (
                int.from_bytes(func(value.encode()).digest(), "little") % self.size
            res |= 2**position
        return res

    def format_hash(self, value: str) -> str:
        return self.format_bin(self.hash_(value))

    def estimated_error_rate(self) -> float:
        n_ones = bin(self.bitarray).count("1")
        return (n_ones / self.size) ** len(HASH_FUNCTIONS)
About this Algorithm

Bloom Filters are one of a class of probabilistic data structures. The Bloom Filter uses hashes and probability to determine whether a particular item is present in a set. It can do so in constant time: O(1) and sub-linear space, though technically still O(n). An important feature of a Bloom Filter is that it is guaranteed never to provide a false negative, saying an element isn't present when it is. However, it has a probability (based on the tuning of its parameters) of providing a false positive, saying an element is present when it is not. The Bloom Filter uses a multi-hash scheme. On insertion, the inserted object is run through each hash, which produces a slot number. That slot number is flipped to 1 in the bit array. During a presence check, the object is run through the same set of hashes, and if each corresponding slot is 1, the filter reports the object has been added. If any of them are 0, it reports that the object has not been added. The hashes must be deterministic and uniformly distributed over the slots for the Bloom filter to operate effectively.


Operation Average
Initialize O(1)
Insertion O(1)
Query O(1)
Space O(n)



  1. Bloom Filter is Initialized, with a number of hash functions that will be run against it (henceforth known as k), and with an array of bits of size M with each bit set to 0. There are 3 distinct schemes to tune these parameters.
    1. M and k are explicitly set by the user
    2. k and M are calculated based off the expected number of elements to minimize false positives.
    3. k and M are calculated based off a desired error rate.


  1. Object is run through k hashes
  2. For each result of the hash n determine the slot within the filter m by calculating n % M = m
  3. Set slot m within the filter to 1


  1. Object is run through k hashes
  2. For each result of the hash n determine the slot within the filter m by calculating n % M = m
  3. Check slot m, if m is set to 0 return false
  4. Return true



As an example, let us look at a Bloom Filter of Strings, we will initialize the Bloom Filter with 10 slots an we will use 3 hashes

slot 0 1 2 3 4 5 6 7 8 9
state 0 0 0 0 0 0 0 0 0 0


Let's try to insert foo, we will run foo through our three hash functions

h1(foo) = 2
h2(foo) = 5
h3(foo) = 6

With hashes run, we will flip the corresponding bits to 1

slot 0 1 2 3 4 5 6 7 8 9
state 0 0 1 0 0 1 1 0 0 0


Query bar

Let's first try querying bar, to query bar we run bar through our three hash functions:

h1(bar) = 3
h2(bar) = 4
h3(bar) = 6

If we look at our bit array, bits 3 and 4 are both not set, if even just 1 bit is not set, we return false, so in this case we return false. bar has not been added

Query foo

Let's now try to query foo, when we run foo through our hashes we get:

h1(foo) = 2
h2(foo) = 5
h3(foo) = 6

Of course, since we already inserted foo, our table has each of the three bits our hashes produced set to 1, so we return true, foo is present

False Positive

Let's say we inserted bar and the current state of our table is:

slot 0 1 2 3 4 5 6 7 8 9
state 0 0 1 1 1 1 1 0 0 0

Let's now query baz, when we run baz through our hash functions we get:

h1(baz) = 3
h2(baz) = 5
h3(baz) = 6

Notice that this does not match either the result of foo or bar, however because slots 3, 5, and 6 are already set, we report true, that baz is in the set, and therefore produce a false positive.

Advantage Over HashSets

  • Significantly more space-efficient, Both are technically O(n) space complexity, but since bloom filters will only take up several bits per item, hash sets must hold the entire item.
  • Presence checks are guaranteed to be O(1) for Bloom Filters, for HashSets, the average is O(1), but worst case is O(n)

Disadvantage v.s. Hash Sets

  • Bloom Filters can report false positives. Optimally there should be about a 1% false-positive rate.
  • Bloom Filters do not store the objects inserted into it, so you cannot recover items inserted.


The probability of false positives increases with the probability of hash collisions within the filter. However, you can optimize the number of collisions if you have some sense of the cardinality of your set ahead of time. You can do this by optimizing k and M, M should be ~ 8-10 bits per expected item, and k should be (M/n) * ln2.

Video Explainer

Video Explainer by Narendra L