1 / 22

Optimal Space Lower Bounds for All Frequency Moments

Optimal Space Lower Bounds for All Frequency Moments. David Woodruff MIT dpwood@mit.edu. 4. 3. 7. 3. 1. 1. 0. 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

lynch
Télécharger la présentation

Optimal Space Lower Bounds for All Frequency Moments

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Optimal Space Lower Bounds for All Frequency Moments David Woodruff MIT dpwood@mit.edu

  2. 4 3 7 3 1 1 0 The Streaming Model … • Stream of elements a1, …, aq 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

  3. Frequency Moments • q = stream size, m = universe size • fi = # 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

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

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

  6. Previous Work • Upper Bounds: Can -approximate F0 [BJKST02],F2 [AMS96], Fk [CK04], k > 2 with space respectively: • Lower Bounds: • [AMS96] 8 k, –approximating Fk need (log m) space • [IW03] -approximating F0 requires space if • Questions: Does the bound hold for k  0? Does it hold for F0 for smaller ?

  7. First Result • Optimal Lower Bound: 8 k  1 and any  = (m-1/2), any -approximator for Fk must use (-2) bits of space. • F1 = q computed trivially in log q space • Fk computed in O(m log q) space, so need  = (m-.5) • Technique: Reduction from 2-party protocol for computing Hamming distance (x,y)

  8. Idea Behind Lower Bounds Alice Bob y 2 {0,1}m x 2 {0,1}m Stream s(y) Stream s(x) S Internal state of A (1 §) Fk algorithm A (1 §) Fk algorithm A • Compute (1 §) Fk(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(,)

  9. 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)?

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

  11. Shatter Coefficient Theorem • Notation: For f: X£Y! {0,1}, define: fX = { fx(y) : Y ! {0,1} | x 2X }, where fx(y) = f(x,y) • Theorem [BJKS]: For every f: X £ Y ! {0,1}, every p ¸ VCD( fX ), R1/3(f) = (log(SC(fX, p)))

  12. Hamming Distance Decision Problem (HDDP) Set t = (1/2) Alice Bob x 2 {0,1}t y 2 {0,1}t • Promise Problem : • (x,y) · t/2 – t1/2 (x,y) > t/2 • f(x,y) = 0 OR f(x,y) = 1 • We will lower bound R1/3(f) via SC(fX, t), but first, a critical lemma…

  13. Main Lemma S µ{0,1}n • 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 • 8 t 2 T, (y, t) · n/2 – cn1/2 for some c > 0 • 8 t 2 S – T, (y,t) > n/2 = T y = S-T

  14. Lemma Solves HDDP Complexity • Theorem: R1/3(f) = (t) = (-2). • Proof: • Alice gets yT for random good set T applying main lemma with n = t. • Bob gets random s 2 S • Let f: {yT }T£ S ! {0,1}. • Main Lemma =>SC(f) = 2(t) • [BJKS] => R1/3(f) = (t) = (-2)

  15. Back to Frequency Moments Idea: Use -approximator for Fk in a protocol to solve HDDP y2 {0,1}t s 2 S µ {0,1}t ith universe element included exactly once in auxiliary stream ay (resp. as) if and only if yi (resp. si) = 1. ay as Fk Alg Fk Alg State

  16. Solving HDDP with Fk • Alice/Bob compute -approx to Fk(ay± as) • Fk(ay± as) = 2k wt(y Æ s) + 1k(y,s) For k  1, • Alice also transmits wt(y) in log m space. Conclusion: -approximating Fk(ay± as) decides HDDP, so space for Fk is (t) = (-2)

  17. 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 • 8 t 2 T, (y, t) · n/2 – cn1/2 for some c > 0 • 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

  18. Proving the Main Lemma Let T ={t1, …, tn/2} µ S be arbitrary Let yi = majority(t1,i, ..., tn/2,i) for all i 2 [m] What is probability p that both: • 8 t 2 T, (y, t) · n/2 – cn1/2 for some c > 0 • 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…

  19. The Matrix Problem • WLOG, assume y = 1n (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 -> … tn/2 -> 101001000101111001 100101011100011110 001110111101010101 101010111011100011 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?

  20. 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?

  21. Second Result • Bipartite graph count: Probabilistic argument shows at least 2n^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 > 2mn-zm-n. • Previous known count: 2mn-m-n [MW – personal comm.] • Follows easily from a correlation inequality of Kleitman. • Our proof uses correlation inequalities, but more involved analysis.

  22. Summary • Results: • Optimal Lower Bound: 8 k  1 and any  = (m-1/2), any -approximator for Fk 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 2mn-zm-n for constant z < 1.

More Related