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Unformatted text preview: 1 More StreamMining Counting Distinct Elements Computing “Moments” Frequent Itemsets Elephants and Troops Exponentially Decaying Windows 2 Counting Distinct Elements ◆ Problem : a data stream consists of elements chosen from a set of size n . Maintain a count of the number of distinct elements seen so far. ◆ Obvious approach : maintain the set of elements seen. 3 Applications ◆ How many different words are found among the Web pages being crawled at a site? ◗ Unusually low or high numbers could indicate artificial pages (spam?). ◆ How many different Web pages does each customer request in a week? 4 Using Small Storage ◆ Real Problem : what if we do not have space to store the complete set? ◆ Estimate the count in an unbiased way. ◆ Accept that the count may be in error, but limit the probability that the error is large. 5 FlajoletMartin* Approach ◆ Pick a hash function h that maps each of the n elements to at least log 2 n bits. ◆ For each stream element a , let r ( a ) be the number of trailing 0’s in h ( a ). ◆ Record R = the maximum r ( a ) seen. ◆ Estimate = 2 R . * Really based on a variant due to AMS (Alon, Matias, and Szegedy) 6 Why It Works ◆ The probability that a given h ( a ) ends in at least r 0’s is 2r . ◆ If there are m different elements, the probability that R ≥ r is 1 – (1  2r ) m . Prob. a given h(a) ends in fewer than r 0’s. Prob. all h(a)’s end in fewer than r 0’s. 7 Why It Works – (2) ◆ Since 2r is small, 1  (12r ) m ≈ 1  e m2 . ◆ If 2 r >> m , 1  (1  2r ) m ≈ 1  (1  m2r ) ≈ m /2 r ≈ 0. ◆ If 2 r << m , 1  (1  2r ) m ≈ 1  e m2 ≈ 1. ◆ Thus, 2 R will almost always be around m .rr First 2 terms of the Taylor expansion of e 8 Why It Doesn’t Work ◆ E(2 R ) is actually infinite. ◗ Probability halves when R > R +1, but value doubles. ◆ Workaround involves using many hash functions and getting many samples. ◆ How are samples combined?...
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 Fall '09
 stream, Frequent Itemsets, exponentially decaying windows

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