Communication Complexity

Transcription

1 Communication Complexity Jaikumar Radhakrishnan School of Technology and Computer Science Tata Institute of Fundamental Research Mumbai 31 May 2012 J 31 May / 13

2 Plan 1 Examples, the model, the set disjointness problem 2 Lower bounds for set disjointness, application to streaming 3 Round elimination, lower bounds for data structure problems 4 Remote generation of random variables, correlated sampling J 31 May / 13

3 The static dictionary problem Universe: U = [M], M = 2 m Set: S U, S = n Space: s cells each capable of holding an element of the universe. Preprocessing: On receiving S, represent it in memory using s cells. Membership queries: Given x U, if x S return some information associated with S. Probes: The queries must be answered using t probes to the memory. 31 May / 13

4 The static dictionary problem Universe: U = [M], M = 2 m Set: S U, S = n Space: s cells each capable of holding an element of the universe. Preprocessing: On receiving S, represent it in memory using s cells. Membership queries: Given x U, if x S return some information associated with S. Probes: The queries must be answered using t probes to the memory. Standard solution Use a sorted table, and answer queries using binary search. Space: n cells, each m bits long. Probes: log n. 31 May / 13

5 The static dictionary problem Universe: U = [M], M = 2 m Set: S U, S = n Space: s cells each capable of holding an element of the universe. Preprocessing: On receiving S, represent it in memory using s cells. Membership queries: Given x U, if x S return some information associated with S. Probes: The queries must be answered using t probes to the memory. Standard solution Use a sorted table, and answer queries using binary search. Space: n cells, each m bits long. Probes: log n. Can we do better? 31 May / 13

6 The Fredman-Komlós-Szemerédi scheme Theorem There exists a solution to the dictionary problem with space s = O(n) and t = O(1). 31 May / 13

7 The Fredman-Komlós-Szemerédi scheme Theorem There exists a solution to the dictionary problem with space s = O(n) and t = O(1). Idea Hash Rehash 31 May / 13

8 The predecession problem Beame and Fich 02 Space: s = O(n log m) cells Pred S (x) = max{y S : y x}. Probes: t = O(min{ log m log log m, log n log log n }) 31 May / 13

9 The predecession problem Beame and Fich 02 Space: s = O(n log m) cells Pred S (x) = max{y S : y x}. Probes: t = O(min{ log m log log m, log n log log n }) Lower bounds Beame and Fich 02: The upper bound is tight for deterministic schemes. Pranab Sen 03: The upper bound is tight for randomized schemes. 31 May / 13

10 Today Upper bound Space: O(nm) cells. Probes: O(log m). Lower bound Even if allowed space polynomial in n, m, and word size polynomial m, there is no scheme that can answer predecessor queries using a constant number of probes. 31 May / 13

11 Van Emde Boas Trees Universe: It is a tree of depth m Subset: A subtree of the big tree Idea: Use the FKS scheme to find the ancestor in the tree where the predecessor is available. J 31 May / 13

12 Data structure to communication protocol Data structure Space: s = 2 a = poly(n) (we will take a (log m) 2 ) Word size: b = poly(m) Probes: t 31 May / 13

13 Data structure to communication protocol Data structure Space: s = 2 a = poly(n) (we will take a (log m) 2 ) Word size: b = poly(m) Probes: t Communication protocol Alice gets x U, Bob gets S U. Determine if the number of elements before x in S is even or odd. In each round, Alice sends a bits, where a = log s. In each round, Bob sends b bits. Alice starts and there are 2t rounds. Alice knows the answer in the end. Error at most ɛ. We call this a (2t, a, b) A ɛ protocol for the problem of size (m, n). 31 May / 13

14 Data structure to communication protocol Data structure Space: s = 2 a = poly(n) (we will take a (log m) 2 ) Word size: b = poly(m) Probes: t Communication protocol Alice gets x U, Bob gets S U. Determine if the number of elements before x in S is even or odd. In each round, Alice sends a bits, where a = log s. In each round, Bob sends b bits. Alice starts and there are 2t rounds. Alice knows the answer in the end. Error at most ɛ. We call this a (2t, a, b) A ɛ protocol for the problem of size (m, n). Goal: if t is constant, this is impossible. 31 May / 13

15 Round elimination Eliminating Alice s message (2t, a, b) A ɛ protocol for problem size (m, n) (2t 1, a, b) ɛ+ 1 protocol for problem size ( m 100t at, n) 2 Eliminating Bob s message (2t 1, a, b) ɛ t (2t 2, a, b) ɛ+ 1 50t protocol for problem size ( m at, n) 2 protocol for problem of size ( m at 2 log t 2 m, n bt 2 ) 31 May / 13

16 End game Eliminate all messages. In the end, m and n are still non-trivial and Alice knows the answer without any messages from Bob! J 31 May / 13

17 Duality Let M be a payoff matrix defined as follows. Error matrix Rows: Each row is indexed by a deterministic protocol π with communication C. Columns: Each column is indexed by an input (x, y). Entries: m π,(x,y) = 1 iff π makes an error on input (x, y). View a randomized protocol as a distribution ρ on the rows of of M. Theorem min max ρ µ ρt Mµ = max min µ ρ ρt Mµ, where ρ is a probability distribution on the rows of M and µ is a probability distribution on the columns of M. 31 May / 13

18 Duality Let M be a payoff matrix defined as follows. Error matrix Rows: Each row is indexed by a deterministic protocol π with communication C. Columns: Each column is indexed by an input (x, y). Entries: m π,(x,y) = 1 iff π makes an error on input (x, y). View a randomized protocol as a distribution ρ on the rows of of M. Theorem min max ρ µ ρt Mµ = max min µ ρ ρt Mµ, where ρ is a probability distribution on the rows of M and µ is a probability distribution on the columns of M. Yao s principle aikumar Lower Radhakrishnan bounds: (School ofto Technology show and Computer that no Science Communication efficient Tata InstituteComplexity of randomized Fundamental Researchprotocol Mumbai) exists, 31 May exhibit a/ 13

19 Average encoding theorem Theorem Let X and Y be random variables with some joint distribution. Let F x be the conditional distribution of Y given X = x. Let F be the marginal distribution of Y. Then, Pr[X = x] F x F 1 (2 ln 2)I[X : Y]. x 31 May / 13

20 Thank you 31 May / 13