Streaming model: Rozdiel medzi revíziami

Verzia zo dňa a času 18:35, 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

Count-Min sketch

• Strict turnstile model
• CM sketch with parameters epsilon and delta
• Array of counters A of depth and width
• 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
• Proof:
  • Let
  • for a fixed row i: , because every other element k has probability 1/w to hash to column h_i(j) and thus it cntributes F[k]/w to
  • 
  • By Markov inequality
  • Probability that this happens in every row i is at most
• 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