A Deterministic Algorithm for Summarizing Asynchronous Streams over a Sliding Window
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1 A Deermnsc Algorhm for Summarzng Asynchronous Sreams over a Sldng ndow Cosas Busch Rensselaer Polyechnc Insue Srkana Trhapura Iowa Sae Unversy
2 Oulne of Talk Inroducon Algorhm Analyss
3 Tme C Daa sream: v v v3 v v 4 5 For smplcy assume un valued elemens 3
4 Mos recen me wndow of duraon C Daa sream: v v v3 v v 4 5 Curren me Goal: Compue he sum of elemens wh me samps n me wndow [ C, C ] C v C 4
5 Example I: All packes on a nework lnk, manan he number of dfferen p sources n he las one hour Example II: Large daabase, connuously manan averages and frequency momens 5
6 Daa sream: v v v3 v v 4 5 Synchronous sream : In ascendng order Asynchronous sream : No order guaraneed 6
7 hy Asynchronous Daa Sreams? Synchronous sream Nework Asynchronous sream Nework delay & mul-pah roung Synchronous Synchronous Asynchronous Merge w/o conrol 7
8 Processng Requremens: One pass processng Small workspace: poly-logarhmc n he sze of daa Fas processng me per elemen Approxmae answers are ok 8
9 Our resuls: A deermnsc daa aggregaon algorhm Tme: O logb log Space: O log B Relave Error: log log S log B X S 9
10 Prevous ork: [Daar, Gons, Indyk, Mowan. SIAM Journal on Compung, 00] Deermnsc, Synchronous Mergng buckes [Trhapura, Xu, Busch, PODC, 006] Randomzed, Asynchronous Random samplng 0
11 Oulne of Talk Inroducon Algorhm Analyss
12 C Daa sream: Tme Curren me For smplcy assume un valued elemens
13 Mos recen me wndow of duraon C Daa sream: Curren me Goal: Compue he sum of elemens wh me samps n me wndow [ C, C ] 3
14 3 4 Dvde me no perods of duraon 4
15 sldng wndow T 3 4 C The sldng wndow may span a mos wo me perods 5
16 sldng wndow S T lef S rgh 3 4 C S S S Sum can be wren as wo sub-sums In wo me perods 6
17 sldng wndow S T Dlef lef S rgh C Drgh 3 4 Daa srucure ha manans an esmae of In lef me perod S lef 7
18 S lef T D lef hou loss of Generaly, Consder daa srucure n me perod [, ] D lef 8
19 Daa srucure consss of varous levels D D lef D D L L s an upper bound of he sum n a perod 9
20 Consder level D Bucke a Level 0 Tme perod Couns up o elemens 0
21 Sream: Increase couner value
22 Sream: Increase couner value
23 Sream: Increase couner value 3
24 Sream: 3... Increase couner value 4
25 Sream: 3... Spl bucke Couner hreshold of reached 5
26 Sream: 3... New buckes have hreshold also 6
27 Sream: 3... Increase approprae bucke 7
28 Sream: 3... Increase approprae bucke 8
29 Sream: Increase approprae bucke 9
30 Sream:... m m x Spl bucke
31 Sream:... m x
32 Sream:... m m m 3 4 x Increase approprae bucke 3
33 Sream:... m m... m x Spl bucke x
34 Sream:... m m... m x x
35 Splng Tree x x k x 4 x x
36 Max deph = Leaf buckes of duraon are no spl any furher log 36
37 Leaf buckes The nal bucke may be spl no many buckes 37
38 Leaf buckes Due o space lmaons we only keep he las buckes a log 38
39 T S Suppose we wan o fnd he sum of elemens n me perod [ T, ] S 39
40 T Consder varous levels of splng hreshold S a a k k a a 40
41 T Frs level wh a leaf bucke ha nersecs melne S a a k k a a 4
42 S T Esmae of S: X x x x z k x x Consder buckes on rgh of melne a xz z a 4
43 OR T Frs level wh a leaf bucke On rgh melne S a a k k a a 43
44 Oulne of Talk Inroducon Algorhm Analyss 44
45 S T Suppose ha we use level n order o compue he esmae 45
46 Sream: k l x b x b r Consder splng hreshold level A daa elemen s couned n he approprae bucke 46
47 Sream: k l k r k l r e can assume ha he elemen s placed n he respecve bucke 47
48 Sream: k l l k l r l r r r e can assume ha when bucke spls he elemen s placed n an arbrary chld bucke 48
49 Sream: k l r l k l r l r r If: l k l r GOOD! Elemen couned n correc bucke 49
50 Sream: k l r l k l r l r r If: l r k r BAD! Elemen couned n wrong bucke 50
51 T Consder Leaf Buckes S k If T k GOOD! 5
52 T Consder Leaf Buckes S k If k T BAD! Elemen couned n wrong bucke 5
53 T Consder Leaf Buckes S k X S Z Z Z :elemens of lef par couned on rgh Z :elemens of rgh par couned on lef 53
54 T k Z elemens of lef par couned on rgh k Mus have been nally nsered n one of hese buckes 54
55 Snce ree deph log Z O ( log ) 55
56 Snce ree deph log Z O ( log ) Smlarly, we can prove Z O ( log ) Therefore: X S Z Z O ( log ) 56
57 Snce a log I can be proven S ( log ) 57
58 Snce a log I can be proven S ( log ) Combned wh X S O ( log ) e oban relave error : X S S 58
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