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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 r 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. r Obvious approach : maintain the set of elements seen. 3 Applications r How many different words are found among the Web pages being crawled at a site? R Unusually low or high numbers could indicate artificial pages (spam?). r How many different Web pages does each customer request in a week? 4 Using Small Storage r Real Problem : what if we do not have space to store the complete set? r Estimate the count in an unbiased way. r Accept that the count may be in error, but limit the probability that the error is large. 5 FlajoletMartin* Approach r Pick a hash function h that maps each of the n elements to at least log 2 bits. r For each stream element a , let r ( ) be the number of trailing 0’s in h ( a ). r Record R = the maximum ( ) seen. r Estimate = 2 . * Really based on a variant due to AMS (Alon, Matias, and Szegedy) 6 Why It Works r The probability that a given h ( a ) ends in at least r 0’s is 2r . r If there are m different elements, the probability that R ≥ is 1 – (1  2) m . Prob. a given h(a) ends in fewer than 0’s. Prob. all h(a)’s end in fewer than 0’s. 7 Why It Works – (2) r Since 2r is small, 1  (12r ) m ≈ 1 e m2 . r If 2 r >> m , 1  (1  2) m ≈ 1  (1  m2r ) ≈ /2 r ≈ 0. r If 2 << , 1  (1  2) m ≈ 1 em2 ≈ 1. r Thus, 2 R will almost always be around .rr First 2 terms of the Taylor expansion of x 8 Why It Doesn’t Work r E(2 R ) is actually infinite. R Probability halves when > +1, but value doubles. r Workaround involves using many hash functions and getting many samples. r How are samples combined? R Average ? What if one very large value? R Median ? All values are a power of 2. 9 Solution r Partition your samples into small groups....
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 Fall '09

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