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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


Streaming model: Rozdiel medzi revíziami

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(Majority element)
(Majority element)
Riadok 25: Riadok 25:
 
***else if(x>0) x--;
 
***else if(x>0) x--;
 
***else { j=k; x=1; }   
 
***else { j=k; x=1; }   
* Correctness:
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* If we are allowed a second pass, we can verify if found j is indeed majority
** If j* is the majority element, at the of of the algorithm it is stored in j and x>0
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** Any element other than j* can be paired with some occurrence of j*. Some occurrences of j* will remain
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* If we are allowed a second pass, we can verify of found j is indeed majority
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This algorithm can be generalized to find all elements with frequency greater than M/k by keeping k-1 counters (Misra and Gries 1982)
 
This algorithm can be generalized to find all elements with frequency greater than M/k by keeping k-1 counters (Misra and Gries 1982)

Verzia zo dňa a času 20:23, 17. apríl 2017

Definition

  • Input: stream of n elements
  • Goal: do one pass through the stream (or only a few passes), use a small memory (e.g. O(1) or O(log n)) and answer a specific query about the stream
    • Good for processing very fast streams, such as IP packets passing a router or measurements of a very low-powered device
    • Not enough memory to store the whole stream, each item should be processed fast
  • Strict turnstile model:
    • underlying set {0,...,n-1}
    • virtual vector F of length n initialized to zeroes
    • stream consists of operations (j,x) meaning F[j]+=x
    • At every point we have F[j]>=0 for each j
  • Cash register model: all values x are positive
  • Example: if all values x are +1, we are simply counting frequencies of elements from {0,...,n-1} on input

Majority element

Boyer–Moore majority vote algorithm 1981

  • Consider the cash register model with x=1 in each update
  • Assume that the stream contains a majority element j, i.e. F[j]>M/2 where M is the sum of all F[k]
  • With this assumption j can be found by a simple algorithm with O(1) memory and O(1) time per element:
    • Keep one candidate element j and a counter x
    • Initialize j to fist element of the stream and x=1
    • For each element k of the stream:
      • if(j==k) x++;
      • else if(x>0) x--;
      • else { j=k; x=1; }
  • If we are allowed a second pass, we can verify if found j is indeed majority

This algorithm can be generalized to find all elements with frequency greater than M/k by keeping k-1 counters (Misra and Gries 1982)

  • If current element has a counter, increase its counter
  • Otherwise if some counter at 0, replace with current element, start at 1
  • Otherwise decrease all counters by 1
  • At the end the final set is guaranteed to contain all elements with frequency more than M/k (but we do not know which they are)
    • Decreasing all counters is equivalent to removing k distinct elements from the input multiset (k-1 counters plus the current element)
    • The counters at the end correspond to up to k-1 distinct items on which operation of removing k distinct elements cannot be applied - all elements with more than M/k occurrences must be among counters

Count-Min sketch

  • Strict turnstile model
  • CM sketch with parameters epsilon and delta
  • Array of counters A of depth d=\lceil \ln(1/\delta )\rceil and width w=\lceil e/\epsilon \rceil
  • Each row i of A has a hash function h_i from {0,...,n-1} to {0,...,w-1} (assume totally random, but a weaker assumption of pairwise independence sufficient for analysis)
  • Update (j,x) does A[i,h_i(j)]+=x for all hash function i=0,...,n-1
  • Query F[j]=? returns min_i A[i,h_i(j)]
  • Let F be the correct answer, F' the answer returned
  • Clearly F'>=F, because each A[i,k] may have contributions from other elements as well
  • With probability at least 1-delta we have F'[j]\leq F[j]+\epsilon \sum _{i}F[i]
  • Proof:
    • Let M=\sum _{k}F[k]
    • for a fixed row i: E[A[i,h_{i}(j)]=F[j]+\sum _{{k\neq j}}F[k]/w, because every other element k has probability 1/w to hash to column h_i(j) and thus it cntributes F[k]/w to
    • E[A[i,h_{i}(j)]\leq F[j]+M\epsilon /e
    • By Markov inequality \Pr(A[i,h_{i}(j)]-F[j]\geq M\epsilon )\leq 1/e
    • Probability that this happens in every row i is at most e^{{-d}}\leq \delta
  • Memory does not depend on n
  • Note: if we insert million elements and then delete all but 4 of them, in the final structure we are very likely to be able to identify the 4 remaining ones as M is only 4

Heavy hitters

  • Consider cash register model, for simplicity let all updates by simple increments by 1 (x=1)
  • Let M = sum of values in F
  • Given parameter phi, output all elements i from {0,...,n-1} such that F[i]>=phi*M (we will solve this problem with approximation)
  • Keep current value of M plus a CM sketch with ln(n/delta) rows (instead of n, one could use an upper bound on M if F is very sparse)
  • Also keep a hash table of elements j with current estimate F'[j]>=phi*M for current M and a min-priority queue (binary heap) of these elements with F'[j] as key (hash table keeps indexes within heap of individual elements)
  • After each update (j,x), get estimated F'[j] and if it is at least phi*M, add/update j in priority queue and hash table
  • Also check if minimum item k in the heap has its key large enough compared to current M, otherwise remove it
    • At each moment at most 1/phi elements in the heap and hash table
  • Memory O(ln(n/delta)/epsilon+1/phi), time for update O(ln(n/delta)+log(1/phi))
  • Theorem: Every item which occurs with count more than phi * M is output, and with probability 1 − delta, no item whose count is less than (phi-epsilon)M is output.
    • Threshold phi*M increases in each step; the only element which may become one of heavy hitters is the one just increased
    • Since count-min sketch gives upper bounds, no heavy hitter is ommitted
    • For each individual element j we have Pr(F'[j]-F[j]>=epsilon*M) <= delta/n
    • By union bound Pr(exists j such that F'[j]-F[j]>=epsilon*M) <= delta
    • So with prob >= 1-delta for every j such that F[j]<(phi-epsilon)*M, we have F'[j]<Phi*M

Sources

  • Presentation at AK Data Science Summit - S. Muthu Muthukrishnan 2013 Count-Min Sketch 10 years later video, slides
  • Cormode G, Muthukrishnan S. An improved data stream summary: the count-min sketch and its applications. Journal of Algorithms. 2005 Apr 1;55(1):58-75. [1]
    • Journal article with count-min sketch
  • Muthukrishnan S. Data streams: Algorithms and applications. Foundations and Trends in Theoretical Computer Science. 2005 Sep 27;1(2):117-236. [2]
    • Survey of streaming model results, count-min sketch starts at page 26
  • Cormode, Graham, and Marios Hadjieleftheriou. "Finding the frequent items in streams of data." Communications of the ACM 52.10 (2009): 97-105. [3]
    • Survey and empirical comparison, include Boyer-Moore as well as count-min sketch
  • Wikipedia Count-min sketch, Streaming algorithms, Boyer-Moore algorithm