Download presentation

Presentation is loading. Please wait.

Published byThomas Baldwin Modified over 4 years ago

1
**An Optimal Algorithm for the Distinct Elements Problem**

Daniel Kane, Jelani Nelson, David Woodruff PODS, 2010

2
Problem Description Given a long stream of values from a universe of size n each value can occur any number of times count the number F0 of distinct values See values one at a time One pass over the stream Too expensive to store set of distinct values Algorithms should: Use a small amount of memory Have fast update time (per value processing time) Have fast recovery time (time to report the answer)

3
**Randomized Approximation Algorithms**

3, 141, 59, 265, 3, 58, 9, 7, 9, 32, 3, , 338, 32, 4, … Consider algorithms that store a subset S of distinct values E.g., S = {3, 9, 32, 265} Main drawback is that S needs to be large to know if next value is a new distinct value Any algorithm (whether it stores a subset of values or not) that is either deterministic or computes F0 exactly must use ¼ F0 memory Hence, algorithms must be randomized and settle for an approximate solution: output F 2 [(1-ε)F0, (1+ε)F0] with good probability

4
**Problem History Long sequence of work on the problem**

Flajolet and Martin introduced problem, FOCS 1983 Alon, Bar-Yossef, Beyer, Brody, Chakrabarti, Durand, Estan, Flajolet, Fisk, Fusy, Gandouet, Gemulla, Gibbons, P. Haas, Indyk, Jayram, Kumar, Martin, Matias, Meunier, Reinwald, Sismanis, Sivakumar, Szegedy, Tirthapura, Trevisan, Varghese, W Previous best algorithm: O(ε-2 log log n + log n) bits of memory and O(ε-2) update and reporting time Known lower bound on the memory: (ε-2 + log n) Our result: Optimal O(ε-2 + log n) bits of memory and O(1) update and reporting time

5
Previous Approaches Suppose we randomly hash F0 values into a hash table of 1/ε2 buckets and keep track of the number C of non-empty buckets If F0 < 1/ε2, there is a way to estimate F0 up to (1 ± ε) from C Problem: if F0 À 1/ε2, with high probability, every bucket contains a value, so there is no information Solution: randomly choose Slog n µ Slog n - 1 µ Slog n - 2 µ S1 µ {1, 2, …, n}, where |Si| ¼ n/2i Problem: It takes 1/ε2 log n bits of memory to keep track of this information stream: 3, 141, 59, 265, 3, 58, 9, 7, 9, 32, 3, , 338, 32, 4, … Si = {1, 3, 7, 9, 265} i-th substream: 3, 265, 3, 9, 7, 9, 3, … Run hashing procedure on each substream There is an i for which the # of distinct values in i-th substream ¼ 1/ε2 Hashing procedure on i-th substream works

6
**Our Techniques Observation: - Have 1/ε2 global buckets**

- In each bucket we keep track of the index i of the set Si for the largest i for which Si contains a value hashed to the bucket - This gives O(1/ε2 log log n) bits of memory New Ideas: - Can show with high probability, at every point in the stream, most buckets contain roughly the same index - We can just keep track of the offsets from this common index - We pack the offsets into machine words and use known fast read/write algorithms to variable length arrays to efficiently update offsets - Occasionally we need to decrement all offsets. Can spread the work across multiple updates

Similar presentations

OK

PODC 20081 Distributed Computation of the Mode Fabian Kuhn Thomas Locher ETH Zurich, Switzerland Stefan Schmid TU Munich, Germany TexPoint fonts used in.

PODC 20081 Distributed Computation of the Mode Fabian Kuhn Thomas Locher ETH Zurich, Switzerland Stefan Schmid TU Munich, Germany TexPoint fonts used in.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Download ppt on multimedia and animation Ppt on tcp ip protocol Ppt on global marketing strategies Ppt on stock market trading Easy ppt on bluetooth Ppt on communication Ppt on steps for success Ppt on ac and dc generator Ppt on land pollution in india Ppt on recycling of wastewater effluent