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Táto stránka sa týka školského roku 2016/17. V školskom roku 2017/18 predmet vyučuje Jakub Kováč, stránku predmetu je https://people.ksp.sk/~kuko/ds


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(Bloom Filter (1970))
(Bloom Filter (Bloom 1970))
Riadok 51: Riadok 51:
 
* thus probability of error <= (1-exp(-nk/m))^k  
 
* thus probability of error <= (1-exp(-nk/m))^k  
  
!check Mitzenmacher & Upfal (2005), pp. 109–111, 308. - less strict independence assumption
+
* In Mitzenmacher & Upfal (2005), pp. 109–111, 308. - less strict independence assumption
  
 
'''Exercise'''  
 
'''Exercise'''  

Verzia zo dňa a času 14:06, 29. apríl 2015

Zdroje

Osnova

  • vlastnosti hešovacích funkcií: totally random, universal, k-wise independent, simple tabulation;
  • chaining, perfect hashing, linear probing,
  • Bloom filters, locality sensitive hashing

V sylabe na skúšku je len perfect hashing, rozoberané na Erikovej prednáške

Bloom Filter (Bloom 1970)

  • supports insert x, test if x is in the set
    • may give false positives, e.g. claim that x is in the set when it is not
    • false negatives do not occur

Algorithm

  • a bit string B[0,...,m-1] of length m, and k hash functions hi : U -> {0, ..., m-1}.
  • insert(x): set bits B[h1(x)], ..., B[hk(x)] to 1.
  • test if x in the set: compute h1(y), ..., hk(y) and check whether all these bits are 1
    • if yes, claim x is in the set, but possibility of error
    • if no, answer no, surely true

Lemma: If hash functions are totally random and independent, the probability of error is at most (1-exp(-nk/m))^k

  • proof later
  • totally random hash functions impractical (need to store hash value for each element of the universe), but the assumption simplifies analysis
  • if we set k = ln(2) m/n, get error (1-\exp(-\ln(2)))^{{\ln(2)m/n}}=2^{{-\ln(2)m/n}}=c^{{-m/n}}, where c=2^{{\ln(2)}}\approx 1.62
  • to get error rate p for some n, we need m=n\log _{c}(1/p)=n\lg(1/p)/\lg(c)=n\lg(1/p)/\ln(2)=n\lg(1/p)\lg(e)
  • for 1% error, we need about m=10n bits of space and k=7
  • memory size and error rate are independent of the size of the universe U
  • compare to a hash table, which needs at least to store data items themselves (e.g. in n lg u bits)
  • if we used k=1 (Bloom filter with one hash function), we need m=n/ln(1/(1-p)) bits, which for p=0.01 is about 99.5n, about 10 times more than with 7 hash functions

Use of Bloom filters

  • e.g. an approximate index of a larger data structure on a disk - if x not in the filter, do not bother looking at the disk, but small number of false positives not a problem
  • Example: Google BigTable maps row label, column label and timestamp to a record, underlies many Google technologies. It uses Bloom filter to check if a combination of row and column label occur in a given fragment of the table
    • For details, see Chang, Fay, et al. "Bigtable: A distributed storage system for structured data." ACM Transactions on Computer Systems (TOCS) 26.2 (2008): 4. [1]
  • see also A. Z. Broder and M. Mitzenmacher. Network applications of Bloom filters: A survey. In Internet Math. Volume 1, Number 4 (2003), 485-509. [2]

Proof of lemma

  • probability that some B[i] is set to 1 by hj(x) is 1/m
  • probability that B[i] is not set to 1 is therefore 1-1/m and since we use k independent hash functions, the probability that B[i] is not set to one by any of them is (1-1/m)^k
  • if we insert n elements, each is hashed by each function independently of other elements (hash functions are random) and thus Pr(B[i]=0)=(1-1/m)^{nk}
  • Pr(B[i]=1) = 1-Pr(B[i]=0)
  • consider a new element y which is not in the set
  • error occurs when B[hj(y)]==1 for j=1..k,
  • this happens with probability Pr(B[i]=1)^k = (1-(1-1/m)^{nk})^k
  • recall that for x>1 we have (1-1/x)^x < 1/e (equality in limit as x->infinity)
  • thus probability of error <= (1-exp(-nk/m))^k
  • In Mitzenmacher & Upfal (2005), pp. 109–111, 308. - less strict independence assumption

Exercise Let us assume that we have separate Bloom filters for sets A and B with the same set k hash functions. How can we create Bloom filters for union and intersection? Will the result be different from filter created directly for the union or for the intersection?

Theory

  • Bloom filter above use about 1.44 n lg(1/p) bits to achieve error rate p. There is lower bound of n lg(1/p) [Carter et al 1978], constant 1.44 can be improved to 1+o(1) using more complex data structures, which are then probably less practical
  • Pagh, Anna, Rasmus Pagh, and S. Srinivasa Rao. "An optimal Bloom filter replacement." SODA 2005. [3]
  • L. Carter, R. Floyd, J. Gill, G. Markowsky, and M. Wegman. Exact and approximate membership testers. STOC 1978, pages 59–65. [4]
  • see also proof of lower bound in Broder and Mitzenmacher 2003 below

Counting Bloom filters

  • support insert and delete
  • each B[i] is a counter with b bits
  • insert increases counters, decrease decreases
  • assume no overflows, or reserve largest value 2^b-1 as infinity, cannot be increased or decreased
  • Fan, Li, et al. "Summary cache: A scalable wide-area web cache sharing protocol." ACM SIGCOMM Computer Communication Review. Vol. 28. No. 4. ACM, 1998. [5]

References

Locality Sensitive Hashing

  • Har-Peled, Sariel, Piotr Indyk, and Rajeev Motwani. "Approximate Nearest Neighbor: Towards Removing the Curse of Dimensionality." Theory of Computing 8.1 (2012): 321-350.
    • Proprocess: A set P of n binary sequences of length k, radius r, gap c
    • Query: binary sequence q of length q
    • Output: If there exist a sequence within Hamming distance r from q, find a sequence within r*c of q with probability at least f.
    • Hash each point in P using L hashing functions, each using O(log n) randomly chosen sequence positions (hidden constant depends on d, r, c), put all results to bins in one table
    • Given q, find it using all L hash function, check every collision, report if distance at most cr. If checked more than 3L collisions, stop.
    • L = O(n^{1/c})
    • Space O(nL), query time O(Ld+L(d/r)log n)
    • Probability of failure less than some constant < 1, can be boosted by repeating independently