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Optimal Space Lower Bounds for All Frequency Moments David Woodruff MIT

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The Streaming Model … Stream of elements a 1, …, a q each in {1, …, m} Want to compute statistics on stream Elements arranged in adversarial order Algorithms given one pass over stream Goal: Minimum space algorithm

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Frequency Moments q = stream size, m = universe size f i = # occurrences of item i Define k-th Frequency Moment: Applications F_0 = # distinct elements in stream, F_1 = q F_2 = repeat rate Compute self-joins in database

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The Best Determininistic Algorithm Trivial Algorithm for F k Store/update frequency f i of each item i Space: m items i, log q bits for each f i Total Space = O(m log q) Negative Result [AMS96]: Any algorithm computing F k exactly must use (m) space. Can we do better?

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Approximating F k Negative Result [AMS96]: Any deterministic algorithm that outputs x with |F k – x| < F k must use (m) space. What about randomized approximation algorithms? Randomized algorithm A -approximates F k if A outputs x with Pr[|F k – x| 2/3

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Previous Work Upper Bounds: Can -approximate F 0 [BJKST02], F 2 [AMS96], F k [CK04], k > 2 with space respectively: Lower Bounds: [AMS96] 8 k, –approximating F k need (log m) space [IW03] -approximating F 0 requires space if Questions: Does the bound hold for k 0? Does it hold for F 0 for smaller ?

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First Result Optimal Lower Bound: 8 k 1 and any = (m -1/2 ), any -approximator for F k must use ( -2 ) bits of space. F 1 = q computed trivially in log q space F k computed in O(m log q) space, so need = (m -.5 ) Technique: Reduction from 2-party protocol for computing Hamming distance (x,y)

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Idea Behind Lower Bounds x 2 {0,1} m y 2 {0,1} m Stream s(x) Stream s(y) (1 § ) F k algorithm A Internal state of A Compute (1 § ) F k (s(x) ± s(y)) w.p. > 2/3 Idea: If can decide f(x,y) w.p. > 2/3, space used by A at least randomized 1-way comm. Complexity of f(,) S AliceBob

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Randomized 1-way comm. complexity Boolean function f: X £ Y ! {0,1} Alice has x 2 X, Bob y 2 Y. Bob wants f(x,y) Only 1 message sent: must be from Alice to Bob Comm. cost of protocol = expected length of longest message sent over all inputs. -error randomized 1-way comm. complexity of f, R (f), is comm. cost of optimal protocol computing f w.p. ¸ 1- How do we lower bound R (f)?

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The VC Dimension [KNR] F = {f : X ! {0,1}} family of Boolean functions f 2 F is length-|X | bitstring For S µ X, shatter coefficient SC(f S ) of S is |{f | S } f 2 F | = # distinct bitstrings when F restricted to S SC(F, p) = max S 2 X, |S| = p SC(f S ) If SC(f S ) = 2 |S|, S shattered by F VC Dimension of F, VCD(F), = size of largest S shattered by F

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Shatter Coefficient Theorem Notation: For f: X £ Y ! {0,1}, define: f X = { f x (y) : Y ! {0,1} | x 2 X }, where f x (y) = f(x,y) Theorem [BJKS]: For every f: X £ Y ! {0,1}, every p ¸ VCD( f X ), R 1/3 (f) = (log(SC(f X, p)))

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Hamming Distance Decision Problem (HDDP) We will lower bound R 1/3 (f) via SC(f X, t), but first, a critical lemma… Set t = (1/ 2 ) x 2 {0,1} t y 2 {0,1} t AliceBob Promise Problem : (x,y) · t/2 – t 1/2 (x,y) > t/2 f(x,y) = 0 OR f(x,y) = 1

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Main Lemma S µ{0,1} n y = T = S-T Show 9 S µ {0,1} n with |S| = n s.t. there exists 2 (n) good sets T µ S so that: 9 a separator y 2 {0,1} n s.t 1. 8 t 2 T, (y, t) · n/2 – cn 1/2 for some c > t 2 S – T, (y,t) > n/2

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Lemma Solves HDDP Complexity Theorem: R 1/3 (f) = (t) = ( -2 ). Proof: 1. Alice gets y T for random good set T applying main lemma with n = t. 2. Bob gets random s 2 S 3. Let f: {y T } T £ S ! {0,1}. 4. Main Lemma =>SC(f) = 2 (t) 5. [BJKS] => R 1/3 (f) = (t) = ( -2 )

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Back to Frequency Moments Idea: Use -approximator for F k in a protocol to solve HDDP y 2 {0,1} t s 2 S µ {0,1} t F k Alg State ayay asas ith universe element included exactly once in auxiliary stream a y (resp. a s ) if and only if y i (resp. s i ) = 1.

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Solving HDDP with F k Alice/Bob compute -approx to F k (a y ± a s ) F k (a y ± a s ) = 2 k wt(y Æ s) + 1 k (y,s) For k 1, Conclusion: -approximating F k (a y ± a s ) decides HDDP, so space for F k is (t) = ( -2 ) Alice also transmits wt(y) in log m space.

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But How to Prove Main Lemma? Recall: show 9 S µ {0,1} n with |S| = n s.t. there exists 2 (n) sets T µ S so that: 9 a separator y 2 {0,1} n s.t 1) 8 t 2 T, (y, t) · n/2 – cn 1/2 for some c > 0 2) 8 t 2 S – T, (y,t) > n/2 Use probabilistic method For S, choose n random elts in {0,1} n Show probability arbitrary T µ S satisfies (1),(2) is > 2 -zn for constant z < 1. Hence expected such T is 2 (n) So exists S with 2 (n) such T Key

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Proving the Main Lemma Let T ={t 1, …, t n/2 } µ S be arbitrary Let y i = majority(t 1,i,..., t n/2,i ) for all i 2 [m] What is probability p that both: 1) 8 t 2 T, (y, t) · n/2 – cn 1/2 for some c > 0 2) 8 t 2 S – T, (y,t) > n/2 For 1, let x = Pr[8 t 2 T, (y,t) · n/2 – cn.5 ] For 2, let y = Pr[8 t 2 S-T, (y,t) > n/2] = 2 -n/2 By independence, p = x ¢ y. It remains to lower bound x…

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The Matrix Problem WLOG, assume y = 1 n (recall y is majority word) Want lower bound Pr[8 t 2 T, (y,t) · n/2 – cn.5 ] Equivalent to matrix problem: t1 -> t2 -> … t n/2 -> Given random n/2 x n binary matrix w/each column majority 1, what is probablity each row has at least n/2 + cn.5 1s?

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Bipartite Graphs Matrix Problem Bipartite Graph Counting Problem: How many bipartite graphs exist on n/2 by n vertices s.t. each left vertex has degree > n/2 + cn.5 and each right vertex degree > n/2? ……

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Second Result Bipartite graph count: Probabilistic argument shows at least 2 n^2/2 – zn/2 –n such bipartite graphs for constant z < 1. Analysis generalizes to show # bipartite graphs on m + n vertices w/each left vertex having degree > n/2 and each right vertex degree > m/2 is > 2 mn-zm-n. Previous known count: 2 mn-m-n [MW – personal comm.] Follows easily from a correlation inequality of Kleitman. Our proof uses correlation inequalities, but more involved analysis.

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Summary Results: Optimal Lower Bound: 8 k 1 and any = (m -1/2 ), any -approximator for F k must use ( -2 ) bits of space. Bipartite Graph Count: # bipartite graphs on m + n vertices w/each left vertex having degree > n/2 and each right vertex degree > m/2 is at least 2 mn-zm-n for constant z < 1.

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