Download presentation

Presentation is loading. Please wait.

Published byMelissa Lyons Modified over 2 years ago

1
Estimating Distinct Elements, Optimally David Woodruff IBM Based on papers with Piotr Indyk, Daniel Kane, and Jelani Nelson

2
Problem Description Given a long string of at most n distinct characters, count the number F 0 of distinct characters See characters one at a time One pass over the string Algorithms must use small memory and fast update time –too expensive to store set of distinct characters –algorithms must be randomized and settle for an approximate solution: output F 2 [(1- ² )F 0, (1+ ² )F 0 ] with, say, good constant probability

3
Algorithm History Flajolet and Martin introduced problem, FOCS 1983 –O(log n) space for fixed ε in random oracle model Alon, Matias and Szegedy –O(log n) space/update time for fixed ε with no oracle Gibbons and Tirthapura –O(ε -2 log n) space and O(ε -2 ) update time Bar-Yossef et al –O(ε -2 log n) space and O(log 1/ε) update time –O(ε -2 log log n + log n) space and O(ε -2 ) update time, essentially –Similar space bound also obtained by Flajolet et al in the random oracle model Kane, Nelson and W –O(ε -2 + log n) space and O(1) update and reporting time –All time complexities are in unit-cost RAM model

4
Lower Bound History Alon, Matias and Szegedy –Any algorithm requires Ω(log n) bits of space Bar-Yossef –Any algorithm requires Ω(ε -1 ) bits of space Indyk and W –If ε > 1/n 1/9, any algorithm needs Ω(ε -2 ) bits of space W –If ε > 1/n 1/2, any algorithm needs Ω(ε -2 ) bits of space Jayram, Kumar and Sivakumar –Simpler proof of Ω(ε -2 ) bound for any ε > 1/m 1/2 Brody and Chakrabarti –Show above lower bounds hold even for multiple passes over the string Combining upper and lower bounds, the complexity of this problem is: Θ(ε -2 + log n) space and Θ(1) update and reporting time

5
Outline for Remainder of Talk Proofs of the Upper Bounds Proofs of the Lower Bounds

6
Hash Functions for Throwing Balls We consider a random mapping f of B balls into C containers and count the number of non-empty containers The expected number of non-empty containers is C – C(1-1/C) B If instead of the mapping f, we use an O(log C/ε)/log log C/ε – wise independent mapping g, then –the expected number of non-empty containers under g is the same as that under f, up to a factor of (1 ± ε) Proof based on approximate inclusion-exclusion –express 1 – (1-1/C) B in terms of a series of binomial coefficients –truncate the series at an appropriate place –use limited independence to handle the remaining terms

7
Fast Hash Functions Use hash functions g that can be evaluated in O(1) time. If g is O(log C/ε)/(log log C/ε)-wise independent, the natural family of polynomial hash functions doesnt work We use theorems due to Pagh, Pagh, and Siegel that construct k-wise independent families for large k, and allow O(1) evaluation time For example, Siegel shows: –Let U = [u] and V = [v] with u = v c for a constant c > 1, and suppose the machine word size is Ω(log v) –Let k = v o(1) be arbitrary –For any constant d > 0 there is a randomized procedure that constructs a k-wise independent hash family H from U to V that succeeds with probability 1-1/v d and requires v d space. Each h 2 H can be evaluated in O(1) time Can show we have sufficiently random hash functions that can be evaluated in O(1) time and represented with O(ε -2 + log n) bits of space

8
Algorithm Outline 1.Set K = 1/ε 2 2.Instantiate a lg n x K bitmatrix A, initializing entries of A to 0 3.Pick random hash functions f: [n]->[n] and g: [n]->[K] 4.Obtain a constant factor approximation R to F 0 somehow 5.Update(i): Set A 1, g(i) = 1, A 2, g(i) = 1, …, A lsb(f(i)), g(i) = 1 6.Estimator: Let T = |{j in [K]: A log (16R/K), j = 1}| Output (32R/K) * ln(1-T/K)/ln(1-1/K)

9
Space Complexity Naively, A is a lg n x K bitmatrix, so O(ε -2 log n) space Better: for each column j, store the identity of the largest row i(j) for which A i, j = 1. Note if A i,j = 1, then A i, j = 1 for all i < i –Takes O(ε -2 log log n) space Better yet: maintain a base level I. For each column j, store max(i(j) – I, 0) –Given an O(1)-approximation R to F 0 at each point in the stream, set I = log R –Dont need to remember i(j) if i(j) < I, since j wont be used in estimator –For the j for which i(j) ¸ I, about 1/2 such j will have i(j) = I, about one fourth such j will have i(j) = I+1, etc. –Total number of bits to store offsets is now only O(K) = O(ε -2 ) with good probability at all points in the stream

10
The Constant Factor Approximation Previous algorithms state that at each point in the stream, with probability 1- δ, the output is an O(1)-approximation to F 0 –The space of such algorithms is O(log n log 1/δ). –Union-bounding over a stream of length m gives O(log n log m) total space We achieve O(log n) space, and guarantee the O(1)-approximation R of the algorithm is non-decreasing –Apply the previous scheme on a log n x log n/(log log n) matrix –For each column, maintain the identity of the deepest row with value 1 –Output 2 i, where i is the largest row containing a constant fraction of 1s –We repeat the procedure O(1) times, and output the median of the estimates –Can show the output is correct with probability 1- O(1/log n) –Then we use the non-decreasing property to union-bound over O(log n) events We only increase the base level every time R increases by a factor of 2 –Note that the base level never decreases

11
Running Time Blandford and Blelloch –Definition: a variable length array (VLA) is a data structure implementing an array C 1, …, C n supporting the following operations: Update(i, x) sets the value of C i to x Read(i) returns C i The C i are allowed to have bit representations of varying lengths len(C i ). –Theorem: there is a VLA using O(n + sum i len(C i )) bits of space supporting worst case O(1) updates and reads, assuming the machine word size is at least log n Store our offsets in a VLA, giving O(1) update time for a fixed base level Occasionally we need to update the base level and decrement offsets by 1 –Show base level only increases after Θ(ε -2 ) updates, so can spread this work across these updates, so O(1) worst-case update time –Copy the data structure, use it for performing this additional work so it doesnt interfere with reporting the correct answer –When base level changes, switch to copy For O(1) reporting time, maintain a count of non-zero containers in a level

12
Outline for Remainder of Talk Proofs of the Upper Bounds Proofs of the Lower Bounds

13
1-Round Communication Complexity Alice:Bob: input xinput y Alice sends a single, randomized message M(x) to Bob Bob outputs g(M(x), y) for a randomized function g g(M(x), y) should equal f(x,y) with constant probability Communication cost CC(f) is |M(x)|, maximized over x and random bits Alice creates a string s(x), runs a randomized algorithm A on s(x), and transmits the state of A(s(x)) to Bob Bob creates a string s(y), continues A on s(y), thus computing A(s(x)s(y)) If A(s(x)s(y)) can be used to solve f(x,y), then space(A) ¸ CC(f) What is f(x,y)?

14
The Ω(log n) Bound Consider equality function: f(x,y) = 1 if and only if x = y for x, y 2 {0,1} n/3 Well known that CC(f) = Ω(log n) for (n/3)-bit strings x and y Let C: {0,1} n/3 -> {0,1} n be an error-correcting code with all codewords of Hamming weight n/10 –If x = y, then C(x) = C(y) –If x != y, then ¢ (C(x), C(y)) = Ω(n) Let s(x) be any string on alphabet size n with i-th character appearing in s(x) if and only if C(x) i = 1. Similarly define s(y) If x = y, then F 0 (s(x)s(y)) = n/10. Else, F 0 (s(x)s(y)) = n/10 + Ω(n) A constant factor approximation to F 0 solves f(x,y)

15
The Ω(ε -2 ) Bound Let r = 1/ε 2. Gap Hamming promise problem for x, y in {0,1} r –f(x,y) = 1 if ¢ (x,y) > 1/(2ε -2 ) –f(x,y) = 0 if ¢ (x,y) < 1/(2ε -2 ) - 1/ε Theorem: CC(f) = Ω(ε -2 ) –Can prove this from the Indexing function –Alice has w 2 {0,1} r, Bob has i in {1, 2, …, r}, output g(w, i) = w i –Well-known that CC(g) = Ω(r) Proof: CC(f) = Ω(r), –Alice sends the seed r of a pseudorandom generator to Bob, so the parties have common random strings z i, …, z r 2 {0,1} r –Alice sets x = coordinate-wise-majority{z i | w j = 1} –Bob sets y = z i –Since the z i are random, if x j = 1, then by properties of majority, with good probability ¢ f(x,y) 1/(2ε -2 ) –Repeat a few times to get concentration

16
The Ω(ε -2 ) Bound Continued Need to create strings s(x) and s(y) to have F 0 (s(x)s(y)) decide whether ¢ (x,y) > 1/(2ε -2 ) or ¢ (x,y) < 1/(2ε -2 ) - 1/ε Let s(x) be a string on n characters where character i appears if and only if x i = 1. Similarly define s(y) F 0 (s(x)s(y)) = (wt(x) + wt(y) + ¢ (x,y))/2 –Alice sends wt(x) to Bob A calculation shows a (1+ε)-approximation to F 0 (s(x)s(y)), together with wt(x) and wt(y), solves the Gap-Hamming problem Total communication is space(A) + log 1/ε = Ω(ε -2 ) It follows that space(A) = Ω(ε -2 )

17
Conclusion Combining upper and lower bounds, the streaming complexity of estimating F 0 up to a (1+ε) factor is: Θ(ε -2 + log n) bits of space and Θ(1) update and reporting time Upper bounds based on careful combination of efficient hashing, sampling and various data structures Lower bounds come from 1-way communication complexity

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google