Sublinear Algorithms for Big Data. Part 4: Random Topics


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1 Sublinear Algorithms for Big Data Part : Random Topics Qin Zhang 11
2 Topic 3: Random sampling in distributed data streams (based on a paper with ormode, Muthukrishnan and Yi, PODS 10, JAM 12) 21
3 Distributed streaming Motivated by database/networking applications Adaptive filters [Olston, Jiang, Widom, SIGMOD 03] A generic geometric approach [Scharfman et al. SIGMOD 06] Prediction models [ormode, Garofalakis, Muthukrishnan, Rastogi, SIGMOD 05] sensor networks network monitoring 31 environment monitoring cloud computing
4 Reservoir sampling [Waterman??; Vitter 85] Maintain a (uniform) sample (w/o replacement) of size s from a stream of n items Every subset of size s has equal probability to be the sample 1
5 Reservoir sampling [Waterman??; Vitter 85] Maintain a (uniform) sample (w/o replacement) of size s from a stream of n items Every subset of size s has equal probability to be the sample When the ith item arrives With probability s/i, use it to replace an item in the current sample chosen uniformly at ranfom With probability 1 s/i, throw it away 2
6 Reservoir sampling from distributed streams When k = 1, reservoir sampling has cost Θ(s log n) When k 2, reservoir sampling has cost O(n) because it s costly to track i S k S 3 S S 1 time
7 Reservoir sampling from distributed streams When k = 1, reservoir sampling has cost Θ(s log n) When k 2, reservoir sampling has cost O(n) because it s costly to track i S k Tracking i approximately? Sampling won t be uniform S 3 S S 1 time
8 Reservoir sampling from distributed streams When k = 1, reservoir sampling has cost Θ(s log n) When k 2, reservoir sampling has cost O(n) because it s costly to track i S k S 3 S 2 Tracking i approximately? Sampling won t be uniform Key observation: We don t have to know the size of the population in order to sample! 53 S 1 time
9 Basic idea: binary Bernoulli sampling 61
10 Basic idea: binary Bernoulli sampling
11 Basic idea: binary Bernoulli sampling onditioned upon a row having s active items, we can draw a sample from the active items 63
12 Basic idea: binary Bernoulli sampling onditioned upon a row having s active items, we can draw a sample from the active items The could maintain a Bernoulli sample of size between s and O(s) 6
13 Random sampling Algorithm [with ormode, Muthu & Yi, PODS 10 JAM 11] Initialize i = 0 sites S 1 S 2 S 3 Sites send in every item w.pr. 2 i In epoch i: S k 171
14 Random sampling Algorithm [with ormode, Muthu & Yi, PODS 10 JAM 11] Initialize i = 0 In epoch i: upper lower Sites send in every item w.pr. 2 i oordinator maintains a lower sample and an upper sample: each received item goes to either with equal prob. sites S 1 S 2 S 3 S k (Each item is included in lower sample w.pr. 2 (i+1) ) 172
15 Random sampling Algorithm [with ormode, Muthu & Yi, PODS 10 JAM 11] Initialize i = 0 In epoch i: upper lower Sites send in every item w.pr. 2 i oordinator maintains a lower sample and an upper sample: each received item goes to either with equal prob. sites S 1 S 2 S 3 S k (Each item is included in lower sample w.pr. 2 (i+1) ) When the lower sample reaches size s, the broadcasts to k sites advance to epoch i i+1 Discards the upper sample Randomlysplitsthelowersampleintoanewlowerandanuppersample 173
16 Random sampling Algorithm [with ormode, Muthu & Yi, PODS 10 JAM 11] Initialize i = 0 In epoch i: upper lower Sites send in every item w.pr. 2 i oordinator maintains a lower sample and an upper sample: each received item goes to either with equal prob. sites S 1 S 2 S 3 S k (Each item is included in lower sample w.pr. 2 (i+1) ) When the lower sample reaches size s, the broadcasts to k sites advance to epoch i i+1 Discards the upper sample Randomlysplitsthelowersampleintoanewlowerandanuppersample orrectness: (1): In epoch i, each item is maintained in w. pr. 2 i 17
17 Random sampling Algorithm [with ormode, Muthu & Yi, PODS 10 JAM 11] Initialize i = 0 In epoch i: upper lower Sites send in every item w.pr. 2 i oordinator maintains a lower sample and an upper sample: each received item goes to either with equal prob. sites S 1 S 2 S 3 S k (Each item is included in lower sample w.pr. 2 (i+1) ) When the lower sample reaches size s, the broadcasts to k sites advance to epoch i i+1 Discards the upper sample Randomlysplitsthelowersampleintoanewlowerandanuppersample orrectness: (1): In epoch i, each item is maintained in w. pr. 2 i (2): Always s items are maintained in 175
18 A running example upper lower Maintain s = 3 samples Epoch 0 (p = 1) sites S 1 S 2 S 3 S 181
19 A running example upper lower Maintain s = 3 samples Epoch 0 (p = 1) sites S 1 S 2 S 3 S
20 A running example upper lower 1 Maintain s = 3 samples Epoch 0 (p = 1) sites S 1 S 2 S 3 S
21 A running example upper lower 1 Maintain s = 3 samples Epoch 0 (p = 1) sites S 1 S 2 S 3 S
22 A running example upper lower 1 2 Maintain s = 3 samples Epoch 0 (p = 1) sites S 1 S 2 S 3 S
23 A running example upper lower Maintain s = 3 samples Epoch 0 (p = 1) sites S 1 S 2 S 3 S
24 A running example upper lower Maintain s = 3 samples Epoch 0 (p = 1) sites S 1 S 2 S 3 S
25 A running example upper lower Maintain s = 3 samples Epoch 0 (p = 1) sites S 1 S 2 S 3 S
26 A running example upper lower Maintain s = 3 samples Epoch 0 (p = 1) Now lower sample = 3 discard upper sample split lower sample advance to Epoch 1 sites S 1 S 2 S 3 S
27 A running example upper lower 1 5 Maintain s = 3 samples Epoch 0 (p = 1) Now lower sample = 3 discard upper sample split lower sample advance to Epoch 1 sites S 1 S 2 S 3 S
28 A running example (cont.) upper lower 1 5 Maintain s = 3 samples Epoch 1 (p = 1/2) sites S 1 S 2 S 3 S
29 A running example (cont.) upper lower 1 5 Maintain s = 3 samples Epoch 1 (p = 1/2) sites S 1 S 2 S 3 S (discard) 192
30 A running example (cont.) upper lower Maintain s = 3 samples Epoch 1 (p = 1/2) sites S 1 S 2 S 3 S (discard)
31 A running example (cont.) upper lower Maintain s = 3 samples Epoch 1 (p = 1/2) sites S 1 S 2 S 3 S (discard)
32 A running example (cont.) upper lower Maintain s = 3 samples Epoch 1 (p = 1/2) sites S 1 S 2 S 3 S (discard) (discard) 195
33 A running example (cont.) upper lower Maintain s = 3 samples Epoch 1 (p = 1/2) sites S 1 S 2 S 3 S (discard) (discard)
34 A running example (cont.) upper lower Maintain s = 3 samples Epoch 1 (p = 1/2) sites S 1 S 2 S 3 S (discard) (discard)
35 A running example (cont.) upper lower Maintain s = 3 samples Epoch 1 (p = 1/2) Again lower sample = 3 discard upper sample split lower sample advance to Epoch 2 sites S 1 S 2 S 3 S (discard) (discard)
36 A running example (cont.) upper lower Maintain s = 3 samples Epoch 1 (p = 1/2) Again lower sample = 3 discard upper sample split lower sample advance to Epoch 2 sites S 1 S 2 S 3 S (discard) (discard)
37 A running example (cont.) upper lower Maintain s = 3 samples Epoch 2 (p = 1/) More items will be discarded locally sites S 1 S 2 S 3 S (discard) (discard)
38 A running example (cont.) upper lower Maintain s = 3 samples Epoch 2 (p = 1/) More items will be discarded locally Intuition: maintain a sample prob. at each site p s/n (n: total # items) without knowing n. sites S 1 S 2 S 3 S (discard) (discard)
39 Random sampling Analysis Initialize i = 0 In epoch i: upper lower Sites send in every item w.pr. 2 i oordinator maintains a lower sample and an upper sample: each received item goes to either with equal prob. sites S 1 S 2 S 3 S k (Each item is included in lower sample w.pr. 2 (i+1) ) When the lower sample reaches size s, the broadcasts to k sites advance to epoch i i+1 Discards the upper sample Splitsthelowersampleintoanewlowersampleandanuppersample Analysis: (Msgs sent) Messages sent per epoch O(k +s) # epochs O(logn) = O((k +s)logn) 211
40 Random sampling Analysis and experiments an be 1. improved to Θ(klog k/s n+slogn) and 2. extended to sliding window cases. 221
41 Random sampling Analysis and experiments an be 1. improved to Θ(klog k/s n+slogn) and 2. extended to sliding window cases. Experiments on the real data set from 1998 world cup logs. log 2 (cost) theory 15 log 2 (cost) cost 10 3 practice log 2 s Basic case cost VS sample size n = ,k = log 2 s Timebased sliding cost VS sample size n = ,k = ( log2 Timebased sliding cost VS window size s = 128,k = 128 ) w
42 Random sampling Analysis and experiments an be 1. improved to Θ(klog k/s n+slogn) and 2. extended to sliding window cases. log 2 (cost) theory 15 practice Experiments on the real data set from 1998 world cup logs log 2 s Basic case cost VS sample size n = ,k = 128 total # items n = 7,000,000 # items sent,000 size of sample s = 128 # sites k =
43 Sampling from a (timebased) sliding window sliding window expired windows frozen window current window t 101
44 Sampling from a (timebased) sliding window sliding window t expired windows frozen window current window Sample for sliding window = need new ideas (1) a subsample of the (unexpired) sample of frozen window + (2) a subsample of the sample of current window by ISWoR 102
45 Sampling from a (timebased) sliding window sliding window expired windows frozen window current window Sample for sliding window = need new ideas (1) a subsample of the (unexpired) sample of frozen window + (2) a subsample of the sample of current window by ISWoR (1), (2) may be sampled by different rates. But as long as both have sizes min{s,# live items}, fine. t 103
46 Sampling from a (timebased) sliding window sliding window expired windows frozen window current window Sample for sliding window = need new ideas (1) a subsample of the (unexpired) sample of frozen window + (2) a subsample of the sample of current window by ISWoR (1), (2) may be sampled by different rates. But as long as both have sizes min{s,# live items}, fine. The key issue: how to guarantee both have sizes s? as items in the frozen window are expiring... t 10
47 Sampling from a (timebased) sliding window sliding window expired windows frozen window current window Sample for sliding window = need new ideas (1) a subsample of the (unexpired) sample of frozen window + (2) a subsample of the sample of current window by ISWoR (1), (2) may be sampled by different rates. But as long as both have sizes min{s,# live items}, fine. The key issue: how to guarantee both have sizes s? as items in the frozen window are expiring... Solution: In the frozen window, find a good sample rate such that the sample size s. t 105
48 Dealing with the frozen window sliding window expired windows frozen window current window t Keep all the levels? Need O(w) communication 111
49 Dealing with the frozen window sliding window expired windows frozen window current window t Keep all the levels? Need O(w) communication (s = 2) Keep most recent sampled items in a level until s of them are also sampled at the next level. Total size: O(slogw) 112
50 Dealing with the frozen window sliding window expired windows frozen window current window t Keep all the levels? Need O(w) communication (s = 2) Keep most recent sampled items in a level until s of them are also sampled at the next level. Total size: O(slogw) Guaranteed: There is a blue window with s sampled items that covers the unexpired portion of the frozen window 113
51 Dealing with the frozen window: The algorithm (s = 2) Each site builds its own levelsampling structure for the current window until it freezes Needs O(slogw) space and O(1) time per item 121
52 Dealing with the frozen window: The algorithm (s = 2) Each site builds its own levelsampling structure for the current window until it freezes Needs O(slogw) space and O(1) time per item When the current window freezes For each level, do a kway merge to build the level of the global structure at the. Total communication O((k + s) log w) 122
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