Data Streams A Tutorial
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1 Data Streams A Tutorial Nicole Schweikardt Goethe-Universität Frankfurt am Main DEIS 10: GI-Dagstuhl Seminar on Data Exchange, Integration, and Streams Schloss Dagstuhl, November 8, 2010
2 Data Streams Situation: massive amounts of data generated automatically continuous, rapid updates Examples: meteorological data (sensor networks) astronomical data network monitoring banking and credit transactions Challenges: cannot wait with processing until all the data has arrived process data on-the-fly cannot afford to store all the data store a sketch data may arrive so rapidly that you cannot even afford to look at each incoming data item sampling NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 2/55
3 Example: Network Monitoring Let A be a node in the world wide web. As input, A receives a stream of packets p 1, p 2, p 3, p 4,..., p m. Each packet p i contains information on the sender s IP address, the destination s IP address, the data that is transmitted Question: How many distinct IP addresses have sent at least one packet through node A? I.e., what is the 0-th frequency moment F 0 of the input stream? Problem: A does not want to store the entire stream p 1, p 2, p 3,..., p m. Solution: A suitable randomised algorithm that computes a good approximate answer: NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 3/55
4 Example: Network Monitoring Let A be a node in the world wide web. As input, A receives a stream of packets p 1, p 2, p 3, p 4,..., p m. Each packet p i contains information on the sender s IP address, the destination s IP address, the data that is transmitted Question: How many distinct IP addresses have sent at least one packet through node A? I.e., what is the 0-th frequency moment F 0 of the input stream? Problem: A does not want to store the entire stream p 1, p 2, p 3,..., p m. Solution: A suitable randomised algorithm that computes a good approximate answer: NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 3/55
5 COMPUTING F 0 Tight bound for approximating F 0 Input: A sequence p 1, p 2, p 3,..., p m of elements in {1,.., n}. Task: Compute the number F 0 of distinct elements in the input. Theorem: (a) Upper Bound: (Flajolet, Martin, FOCS 83) For every c > 2 there is a randomized one-pass algorithm that uses O(log n) bits of memory and computes a number Y such that Prob ( Y 1 or Y c ) 2/c. F 0 c F 0 (b) Lower Bound: (Alon, Matias, Szegedy, STOC 96) Any randomized one-pass algorithm computing a number Y such that Prob ( Y Y 0.9 or 1.1 ) 0.25 uses Ω(log n) bits of memory. F 0 F 0 Remark: improved bounds: Bar-Yossef, Jayram, Kumar, Sivakumar (RANDOM 00) and Kane, Nelson, Woodruff (PODS 10). Main issues concerning data streams: How to design algorithms & how to prove lower bounds NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 4/55
6 COMPUTING F 0 Tight bound for approximating F 0 Input: A sequence p 1, p 2, p 3,..., p m of elements in {1,.., n}. Task: Compute the number F 0 of distinct elements in the input. Theorem: (a) Upper Bound: (Flajolet, Martin, FOCS 83) For every c > 2 there is a randomized one-pass algorithm that uses O(log n) bits of memory and computes a number Y such that Prob ( Y 1 or Y c ) 2/c. F 0 c F 0 (b) Lower Bound: (Alon, Matias, Szegedy, STOC 96) Any randomized one-pass algorithm computing a number Y such that Prob ( Y Y 0.9 or 1.1 ) 0.25 uses Ω(log n) bits of memory. F 0 F 0 Remark: improved bounds: Bar-Yossef, Jayram, Kumar, Sivakumar (RANDOM 00) and Kane, Nelson, Woodruff (PODS 10). Main issues concerning data streams: How to design algorithms & how to prove lower bounds NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 4/55
7 COMPUTING F 0 Tight bound for approximating F 0 Input: A sequence p 1, p 2, p 3,..., p m of elements in {1,.., n}. Task: Compute the number F 0 of distinct elements in the input. Theorem: (a) Upper Bound: (Flajolet, Martin, FOCS 83) For every c > 2 there is a randomized one-pass algorithm that uses O(log n) bits of memory and computes a number Y such that Prob ( Y 1 or Y c ) 2/c. F 0 c F 0 (b) Lower Bound: (Alon, Matias, Szegedy, STOC 96) Any randomized one-pass algorithm computing a number Y such that Prob ( Y Y 0.9 or 1.1 ) 0.25 uses Ω(log n) bits of memory. F 0 F 0 Remark: improved bounds: Bar-Yossef, Jayram, Kumar, Sivakumar (RANDOM 00) and Kane, Nelson, Woodruff (PODS 10). Main issues concerning data streams: How to design algorithms & how to prove lower bounds NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 4/55
8 Overview One pass over a single stream Several passes over a single stream Several passes over several streams in parallel Read/write streams Future tasks NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 5/55
9 Overview One pass over a single stream Several passes over a single stream Several passes over several streams in parallel Read/write streams Future tasks NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 6/55
10 One pass over a single stream Scenario: input: memory buffer NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 7/55
11 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
12 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
13 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
14 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
15 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
16 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
17 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
18 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
19 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
20 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
21 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
22 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n Clever Solution: Store running sum O(log n) bits suffice s := x 1 + x 2 + x 3 + x x n 1 Missing number = n (n+1) 2 s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
23 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n Clever Solution: Store running sum O(log n) bits suffice s := x 1 + x 2 + x 3 + x x n 1 Missing number = n (n+1) 2 s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
24 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n Clever Solution: Store running sum O(log n) bits suffice s := x 1 + x 2 + x 3 + x x n 1 Missing number = n (n+1) 2 s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
25 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n Clever Solution: Store running sum O(log n) bits suffice s := x 1 + x 2 + x 3 + x x n 1 Missing number = n (n+1) 2 s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
26 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n Clever Solution: Store running sum O(log n) bits suffice s := x 1 + x 2 + x 3 + x x n 1 Missing number = n (n+1) 2 s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
27 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n Clever Solution: Store running sum O(log n) bits suffice s := x 1 + x 2 + x 3 + x x n 1 Missing number = n (n+1) 2 s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
28 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n Clever Solution: Store running sum O(log n) bits suffice s := x 1 + x 2 + x 3 + x x n 1 Missing number = n (n+1) 2 s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
29 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n Clever Solution: Store running sum O(log n) bits suffice s := x 1 + x 2 + x 3 + x x n 1 Missing number = n (n+1) 2 s Lower Bound: NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
30 MISSING NUMBER Missing Number Puzzle Input: A stream x 1, x 2, x 3,.., x n 1 of n 1 distinct numbers from {1,.., n}. Question: Which number from {1,.., n} is missing? Naive Solution: n requires n bits of storage n Clever Solution: Store running sum O(log n) bits suffice s := x 1 + x 2 + x 3 + x x n 1 Missing number = n (n+1) 2 s Lower Bound: at least log n bits are necessary NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 8/55
31 Exercise # 1 Find a data stream algorithm that uses at most poly(k log n) bits of memory and solves the following generalization of the missing numbers puzzle : k MISSING NUMBERS Input: Two numbers n, k and a stream x 1, x 2, x 3,.., x n k of n k distinct numbers from {1,.., n} Task: Find the k missing numbers NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 9/55
32 The MULTISET-EQUALITY Problem (1/3) MULTISET-EQUALITY Input: Two multisets {x 1,.., x m} and {y 1,.., y m} of numbers x i, y j in {1,.., n}. Question: Is {x 1,.., x m} = {y 1,.., y m}? Total input length: N = O(m log n) bits Observation: Every deterministic solution requires Ω(N) bits of storage. Proof: Use fact from Communication Complexity: NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 10/55
33 Communication Complexity Yao s 2-Party Communication Model: 2 players: Alice & Bob both know a function f : A B {0, 1} Alice only sees input a A, Bob only sees input b B they jointly want to compute f (a, b) Goal: exchange as few bits of communication as possible Fact: Deciding if two m-element input sets a = {x 1,.., x m} {1,.., n} und b = {y 1,.., y m} {1,.., n} are equal, requires at least log ( n m) bits of communication. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 11/55
34 Communication Complexity Yao s 2-Party Communication Model: 2 players: Alice & Bob both know a function f : A B {0, 1} Alice only sees input a A, Bob only sees input b B they jointly want to compute f (a, b) Goal: exchange as few bits of communication as possible Fact: Deciding if two m-element input sets a = {x 1,.., x m} {1,.., n} und b = {y 1,.., y m} {1,.., n} are equal, requires at least log ( n m) bits of communication. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 11/55
35 The MULTISET-EQUALITY Problem (1/3) MULTISET-EQUALITY Input: Two multisets {x 1,.., x m} and {y 1,.., y m} of numbers x i, y j in {1,.., n}. Question: Is {x 1,.., x m} = {y 1,.., y m}? Total input length: N = O(m log n) bits Observation: Every deterministic solution requires Ω(N) bits of storage. Proof: Use fact from Communication Complexity: NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 12/55
36 The MULTISET-EQUALITY Problem (1/3) MULTISET-EQUALITY Input: Two multisets {x 1,.., x m} and {y 1,.., y m} of numbers x i, y j in {1,.., n}. Question: Is {x 1,.., x m} = {y 1,.., y m}? Total input length: N = O(m log n) bits Observation: Every deterministic solution requires Ω(N) bits of storage. Proof: Use fact from Communication Complexity: Deciding if two m-element subsets of {1,.., n} are equal requires at least log ( n m) bits of communication. If n = m 2, then log ( n m) m log m bits of communication are necessary, and the total length of the corresponding MULTISET-EQUALITY input is N = Θ(m log m). NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 12/55
37 The MULTISET-EQUALITY Problem (1/3) MULTISET-EQUALITY Input: Two multisets {x 1,.., x m} and {y 1,.., y m} of numbers x i, y j in {1,.., n}. Question: Is {x 1,.., x m} = {y 1,.., y m}? Total input length: N = O(m log n) bits Observation: Every deterministic solution requires Ω(N) bits of storage. Proof: Use fact from Communication Complexity: Deciding if two m-element subsets of {1,.., n} are equal requires at least log ( n m) bits of communication. If n = m 2, then log ( n m) m log m bits of communication are necessary, and the total length of the corresponding MULTISET-EQUALITY input is N = Θ(m log m). NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 12/55
38 The MULTISET-EQUALITY Problem (2/3) Proof (continued): Known: N = Θ(m log m), and m log m bits of communication are necessary for solving MULTISET-EQUALITY. A deterministic data stream algorithm solving MULTISET-EQUALITY with s bits of storage would lead to a communication protocol with s bits of communication. Thus: Lower bound on communication complexity lower bound on memory size of data stream algorithm NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 13/55
39 The MULTISET-EQUALITY Problem (2/3) Proof (continued): Known: N = Θ(m log m), and m log m bits of communication are necessary for solving MULTISET-EQUALITY. A deterministic data stream algorithm solving MULTISET-EQUALITY with s bits of storage would lead to a communication protocol with s bits of communication. Thus: Lower bound on communication complexity lower bound on memory size of data stream algorithm NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 13/55
40 The MULTISET-EQUALITY Problem (2/3) Proof (continued): Known: N = Θ(m log m), and m log m bits of communication are necessary for solving MULTISET-EQUALITY. A deterministic data stream algorithm solving MULTISET-EQUALITY with s bits of storage would lead to a communication protocol with s bits of communication. ALICE BOB x 1 x 2 x x m y 1 y 2 y y m data stream algorithm memory buffer Thus: Lower bound on communication complexity lower bound on memory size of data stream algorithm NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 13/55
41 The MULTISET-EQUALITY Problem (2/3) Proof (continued): Known: N = Θ(m log m), and m log m bits of communication are necessary for solving MULTISET-EQUALITY. A deterministic data stream algorithm solving MULTISET-EQUALITY with s bits of storage would lead to a communication protocol with s bits of communication. ALICE BOB x 1 x 2 x x m y 1 y 2 y y m data stream algorithm memory buffer Thus: Lower bound on communication complexity lower bound on memory size of data stream algorithm NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 13/55
42 Theorem: The MULTISET-EQUALITY Problem (3/3) (Grohe, Hernich, S., PODS 06) The MULTISET-EQUALITY problem can be solved by a randomised algorithm using O(log N) bits of storage in the following sense: Given m, n, and a stream of numbers a 1,.., a m, b 1,.., b m from {1,.., n}, the algorithm accepts with probability 1 if {a 1,.., a m} = {b 1,.., b m} rejects with probability 0.9 if {a 1,.., a m} = {b 1,.., b m}. Basic idea: Use Fingerprinting -techniques: represent {a 1,.., a m} by a polynomial f (x) := m i=1 x a i represent {b 1,.., b m} by a polynomial g(x) := m i=1 x b i choose a random number r and check if f (r) = g(r) accept if f (r) = g(r); reject otherwise. If {a 1,.., a m} = {b 1,.., b m}, then f (x) = g(x), and thus the algorithm always accepts. If {a 1,.., a m} = {b 1,.., b m}, then there are at most degree(f g) many distinct r with f (r) = g(r), and thus the algorithm rejects with high probability. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 14/55
43 Theorem: The MULTISET-EQUALITY Problem (3/3) (Grohe, Hernich, S., PODS 06) The MULTISET-EQUALITY problem can be solved by a randomised algorithm using O(log N) bits of storage in the following sense: Given m, n, and a stream of numbers a 1,.., a m, b 1,.., b m from {1,.., n}, the algorithm accepts with probability 1 if {a 1,.., a m} = {b 1,.., b m} rejects with probability 0.9 if {a 1,.., a m} = {b 1,.., b m}. Basic idea: Use Fingerprinting -techniques: represent {a 1,.., a m} by a polynomial f (x) := m i=1 x a i represent {b 1,.., b m} by a polynomial g(x) := m i=1 x b i choose a random number r and check if f (r) = g(r) accept if f (r) = g(r); reject otherwise. If {a 1,.., a m} = {b 1,.., b m}, then f (x) = g(x), and thus the algorithm always accepts. If {a 1,.., a m} = {b 1,.., b m}, then there are at most degree(f g) many distinct r with f (r) = g(r), and thus the algorithm rejects with high probability. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 14/55
44 Theorem: The MULTISET-EQUALITY Problem (3/3) (Grohe, Hernich, S., PODS 06) The MULTISET-EQUALITY problem can be solved by a randomised algorithm using O(log N) bits of storage in the following sense: Given m, n, and a stream of numbers a 1,.., a m, b 1,.., b m from {1,.., n}, the algorithm accepts with probability 1 if {a 1,.., a m} = {b 1,.., b m} rejects with probability 0.9 if {a 1,.., a m} = {b 1,.., b m}. Basic idea: Use Fingerprinting -techniques: represent {a 1,.., a m} by a polynomial f (x) := m i=1 x a i represent {b 1,.., b m} by a polynomial g(x) := m i=1 x b i choose a random number r and check if f (r) = g(r) accept if f (r) = g(r); reject otherwise. If {a 1,.., a m} = {b 1,.., b m}, then f (x) = g(x), and thus the algorithm always accepts. If {a 1,.., a m} = {b 1,.., b m}, then there are at most degree(f g) many distinct r with f (r) = g(r), and thus the algorithm rejects with high probability. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 14/55
45 Theorem: The MULTISET-EQUALITY Problem (3/3) (Grohe, Hernich, S., PODS 06) The MULTISET-EQUALITY problem can be solved by a randomised algorithm using O(log N) bits of storage in the following sense: Given m, n, and a stream of numbers a 1,.., a m, b 1,.., b m from {1,.., n}, the algorithm accepts with probability 1 if {a 1,.., a m} = {b 1,.., b m} rejects with probability 0.9 if {a 1,.., a m} = {b 1,.., b m}. Basic idea: Use Fingerprinting -techniques: represent {a 1,.., a m} by a polynomial f (x) := m i=1 x a i represent {b 1,.., b m} by a polynomial g(x) := m i=1 x b i choose a random number r and check if f (r) = g(r) accept if f (r) = g(r); reject otherwise. If {a 1,.., a m} = {b 1,.., b m}, then f (x) = g(x), and thus the algorithm always accepts. If {a 1,.., a m} = {b 1,.., b m}, then there are at most degree(f g) many distinct r with f (r) = g(r), and thus the algorithm rejects with high probability. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 14/55
46 Theorem: The MULTISET-EQUALITY Problem (3/3) (Grohe, Hernich, S., PODS 06) The MULTISET-EQUALITY problem can be solved by a randomised algorithm using O(log N) bits of storage in the following sense: Given m, n, and a stream of numbers a 1,.., a m, b 1,.., b m from {1,.., n}, the algorithm accepts with probability 1 if {a 1,.., a m} = {b 1,.., b m} rejects with probability 0.9 if {a 1,.., a m} = {b 1,.., b m}. Basic idea: Use Fingerprinting -techniques: represent {a 1,.., a m} by a polynomial f (x) := m i=1 x a i represent {b 1,.., b m} by a polynomial g(x) := m i=1 x b i choose a random number r and check if f (r) = g(r) accept if f (r) = g(r); reject otherwise. If {a 1,.., a m} = {b 1,.., b m}, then f (x) = g(x), and thus the algorithm always accepts. If {a 1,.., a m} = {b 1,.., b m}, then there are at most degree(f g) many distinct r with f (r) = g(r), and thus the algorithm rejects with high probability. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 14/55
47 Exercise # 2 Work out the details of the described algorithm and its analysis. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 15/55
48 Overview One pass over a single stream Several passes over a single stream Several passes over several streams in parallel Read/write streams Future tasks NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 16/55
49 Scenario: input: Several passes over a single stream memory buffer Parameters: p : number of passes s : size of memory buffer (number of bits) We call such computations (p, s)-bounded computations. If necessary, an output stream can be generated during a computation. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 17/55
50 Scenario: input: Several passes over a single stream memory buffer Parameters: p : number of passes s : size of memory buffer (number of bits) We call such computations (p, s)-bounded computations. If necessary, an output stream can be generated during a computation. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 17/55
51 input: An easy observation Fact: memory buffer During a (p, s)-bounded computation, only (p s) bits can be communicated between the first and the second half of the input. Consequence: Lower bounds on communication complexity lead to lower bounds for (p, s)-bounded computations... even if backward passes are allowed... even if writing on the input tape is allowed. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 18/55
52 input: An easy observation Fact: memory buffer During a (p, s)-bounded computation, only (p s) bits can be communicated between the first and the second half of the input. Consequence: Lower bounds on communication complexity lead to lower bounds for (p, s)-bounded computations... even if backward passes are allowed... even if writing on the input tape is allowed. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 18/55
53 input: An easy observation Fact: memory buffer During a (p, s)-bounded computation, only (p s) bits can be communicated between the first and the second half of the input. Consequence: Lower bounds on communication complexity lead to lower bounds for (p, s)-bounded computations... even if backward passes are allowed... even if writing on the input tape is allowed. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 18/55
54 A lower bound for connectedness of a graph CONNECTEDNESS Parameters: m edges on n nodes Input: A list of edges e 1,..., e m on node set V {1,.., n}. Question: Is the input graph connected? Theorem: (Henzinger, Raghavan, Rajagopalan, 1998) Solving CONNECTEDNESS with p passes requires Ω(n/p) bits of memory. Proof: By a reduction using the set disjointness problem. SET DISJOINTNESS PROBLEM Input: Two sets A, B {1,.., n} Question: Is A B =? Known communication complexity of the set disjointness problem: n bits of communication are necessary (and sufficient). NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 19/55
55 A lower bound for connectedness of a graph CONNECTEDNESS Parameters: m edges on n nodes Input: A list of edges e 1,..., e m on node set V {1,.., n}. Question: Is the input graph connected? Theorem: (Henzinger, Raghavan, Rajagopalan, 1998) Solving CONNECTEDNESS with p passes requires Ω(n/p) bits of memory. Proof: By a reduction using the set disjointness problem. SET DISJOINTNESS PROBLEM Input: Two sets A, B {1,.., n} Question: Is A B =? Known communication complexity of the set disjointness problem: n bits of communication are necessary (and sufficient). NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 19/55
56 Exercise # 3 Work out the details of the proof: (a) prove that n bits of communication are necessary for solving the set disjointness problem in Yao s 2-party communication model, and (b) use this to show that solving graph connectedness with p passes requires Ω(n/p) bits of memory. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 20/55
57 A lower bound for sorting SORTING Input length N = O(m log n) bits Input: A sequence of numbers x 1,..., x m {1,.., n} (for arbitrary m, n). Output: x 1,..., x m sorted in ascending order. Theorem: (Grohe, Koch, S., ICALP 05) SORTING can be solved by a (p, s)-bounded computation (p s) Ω(N) Proof: upper bound: easy. lower bound: by a reduction using the set disjointness problem. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 21/55
58 A hierarchy on the number of passes Allowing a single extra scan may be more powerful than significantly increasing the internal memory space: Theorem: (Hernich, S., Theor. Comput. Sci. 2008) For every logspace-computable function p with p(n) o ( ) N log 2 N, there exists a decision problem that can be solved by a (p+1, s)-bounded computation, but that cannot be solved by any (p, S)-bounded computation, for s(n) = O(log N) and S(N) = o ( ) N p(n) log N. Remark: An analogous result also holds for randomised computations. Proof idea: Use a result by Nisan and Wigderson (1993) on the k-round communication complexity of a particular pointer jumping problem. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 22/55
59 A hierarchy on the number of passes Allowing a single extra scan may be more powerful than significantly increasing the internal memory space: Theorem: (Hernich, S., Theor. Comput. Sci. 2008) For every logspace-computable function p with p(n) o ( ) N log 2 N, there exists a decision problem that can be solved by a (p+1, s)-bounded computation, but that cannot be solved by any (p, S)-bounded computation, for s(n) = O(log N) and S(N) = o ( ) N p(n) log N. Remark: An analogous result also holds for randomised computations. Proof idea: Use a result by Nisan and Wigderson (1993) on the k-round communication complexity of a particular pointer jumping problem. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 22/55
60 A hierarchy on the number of passes Allowing a single extra scan may be more powerful than significantly increasing the internal memory space: Theorem: (Hernich, S., Theor. Comput. Sci. 2008) For every logspace-computable function p with p(n) o ( ) N log 2 N, there exists a decision problem that can be solved by a (p+1, s)-bounded computation, but that cannot be solved by any (p, S)-bounded computation, for s(n) = O(log N) and S(N) = o ( ) N p(n) log N. Remark: An analogous result also holds for randomised computations. Proof idea: Use a result by Nisan and Wigderson (1993) on the k-round communication complexity of a particular pointer jumping problem. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 22/55
61 A lower bound for finding a longest increasing subsequence LONGEST-INCREASING-SUBSEQUENCE Input: a sequence of numbers x 1,..., x m {1,.., n} (for arbitrary m, n) Output: an increasing subsequence x i1,..., x ik of maximum length (denoted k) Theorem: (Guha, McGregor, ICALP 08) Any randomized p-pass algorithm solving LONGEST-INCREASING-SUBSEQUENCE with p passes (and probability 0.9) requires Ω ( k p 1 ) bits of memory. Proof: not by using communication complexity introduce a new method of pass elimination (somewhat related to round elimination methods in communication complexity, but taylored towards stream processing). Remark: A matching upper bound was proved by Liben-Nowell, Vee, Zhu, COCOON 05. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 23/55
62 A lower bound for finding a longest increasing subsequence LONGEST-INCREASING-SUBSEQUENCE Input: a sequence of numbers x 1,..., x m {1,.., n} (for arbitrary m, n) Output: an increasing subsequence x i1,..., x ik of maximum length (denoted k) Theorem: (Guha, McGregor, ICALP 08) Any randomized p-pass algorithm solving LONGEST-INCREASING-SUBSEQUENCE with p passes (and probability 0.9) requires Ω ( k p 1 ) bits of memory. Proof: not by using communication complexity introduce a new method of pass elimination (somewhat related to round elimination methods in communication complexity, but taylored towards stream processing). Remark: A matching upper bound was proved by Liben-Nowell, Vee, Zhu, COCOON 05. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 23/55
63 Overview One pass over a single stream Several passes over a single stream Several passes over several streams in parallel Read/write streams Future tasks NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 24/55
64 Several passes over several streams in parallel Basic scenario: input S: input T: memory buffer Parameters: 2 input streams: S = s 1, s 2,..., s n and T = t 1, t 2,..., t n. one pass over each input; heads may proceed asynchronously advancement of heads and new content of memory depends on the current content of memory and the symbols seen at both heads for simplicity: advancement of only one head at a time s : size of memory buffer (number of bits) m : number of possible memory configurations, i.e., log m = s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 25/55
65 Several passes over several streams in parallel Basic scenario: input S: input T: memory buffer Parameters: 2 input streams: S = s 1, s 2,..., s n and T = t 1, t 2,..., t n. one pass over each input; heads may proceed asynchronously advancement of heads and new content of memory depends on the current content of memory and the symbols seen at both heads for simplicity: advancement of only one head at a time s : size of memory buffer (number of bits) m : number of possible memory configurations, i.e., log m = s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 25/55
66 Several passes over several streams in parallel Basic scenario: input S: input T: memory buffer Parameters: 2 input streams: S = s 1, s 2,..., s n and T = t 1, t 2,..., t n. one pass over each input; heads may proceed asynchronously advancement of heads and new content of memory depends on the current content of memory and the symbols seen at both heads for simplicity: advancement of only one head at a time s : size of memory buffer (number of bits) m : number of possible memory configurations, i.e., log m = s NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 25/55
67 How to prove lower bounds in this scenario? Problem: Classical communication complexity results cannot be used so easily here. Solution: Take a direct look at the flow of information during computations. Consider the following example: n 2 D n := {a 1, b 1, c 1..., a n, b n, c n} domain of 3n input items variation of the set disjointness problem: DISJ n Input: Two streams S = s 1, s 2,..., s n and T = t 1, t 2,..., t n of elements in D n Question: Is {s 1, s 2,..., s n} {t 1, t 2,..., t n} =? NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 26/55
68 How to prove lower bounds in this scenario? Problem: Classical communication complexity results cannot be used so easily here. Solution: Take a direct look at the flow of information during computations. Consider the following example: n 2 D n := {a 1, b 1, c 1..., a n, b n, c n} domain of 3n input items variation of the set disjointness problem: DISJ n Input: Two streams S = s 1, s 2,..., s n and T = t 1, t 2,..., t n of elements in D n Question: Is {s 1, s 2,..., s n} {t 1, t 2,..., t n} =? NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 26/55
69 How to prove lower bounds in this scenario? Problem: Classical communication complexity results cannot be used so easily here. Solution: Take a direct look at the flow of information during computations. Consider the following example: n 2 D n := {a 1, b 1, c 1..., a n, b n, c n} domain of 3n input items variation of the set disjointness problem: DISJ n Input: Two streams S = s 1, s 2,..., s n and T = t 1, t 2,..., t n of elements in D n such that s i {a i, b i } and t n i+1 {a i, c i } Question: Is {s 1, s 2,..., s n} {t 1, t 2,..., t n} =? NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 26/55
70 A lower bound proof for DISJ n (1/5) DISJ n D n := {a 1, b 1, c 1..., a n, b n, c n} Input: two streams S = s 1, s 2,..., s n, T = t 1, t 2,..., t n of elements in D n, such that s i {a i, b i } and t n i+1 {a i, c i }. Question: Is {s 1, s 2,..., s n} {t 1, t 2,..., t n} =? Theorem: (Bar Yossef, Shalem, ICDE 08) DISJ n cannot be solved by a deterministic algorithm that performs one pass over each stream and that uses less than n log n 1 bits of memory. Proof: Consider input instances D(I 1, I 2 ) := (S I1, T I2 ) with I 1, I 2 {1,.., n} and S I1 : i I 1 = i-th position carries a i i I 1 = i-th position carries b i T I2 : i I 2 = (n i+1)-th position carries a i i I 2 = (n i+1)-th position carries c i Note: S I1 T I2 = I 1 I 2 = Restrict attention to input instances D(I, I) = (S I, T I ) for I {1,.., n}. (particular yes -instances) NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 27/55
71 A lower bound proof for DISJ n (1/5) DISJ n D n := {a 1, b 1, c 1..., a n, b n, c n} Input: two streams S = s 1, s 2,..., s n, T = t 1, t 2,..., t n of elements in D n, such that s i {a i, b i } and t n i+1 {a i, c i }. Question: Is {s 1, s 2,..., s n} {t 1, t 2,..., t n} =? Theorem: (Bar Yossef, Shalem, ICDE 08) DISJ n cannot be solved by a deterministic algorithm that performs one pass over each stream and that uses less than n log n 1 bits of memory. Proof: Consider input instances D(I 1, I 2 ) := (S I1, T I2 ) with I 1, I 2 {1,.., n} and S I1 : i I 1 = i-th position carries a i i I 1 = i-th position carries b i T I2 : i I 2 = (n i+1)-th position carries a i i I 2 = (n i+1)-th position carries c i Note: S I1 T I2 = I 1 I 2 = Restrict attention to input instances D(I, I) = (S I, T I ) for I {1,.., n}. (particular yes -instances) NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 27/55
72 A lower bound proof for DISJ n (1/5) DISJ n D n := {a 1, b 1, c 1..., a n, b n, c n} Input: two streams S = s 1, s 2,..., s n, T = t 1, t 2,..., t n of elements in D n, such that s i {a i, b i } and t n i+1 {a i, c i }. Question: Is {s 1, s 2,..., s n} {t 1, t 2,..., t n} =? Theorem: (Bar Yossef, Shalem, ICDE 08) DISJ n cannot be solved by a deterministic algorithm that performs one pass over each stream and that uses less than n log n 1 bits of memory. Proof: Consider input instances D(I 1, I 2 ) := (S I1, T I2 ) with I 1, I 2 {1,.., n} and S I1 : i I 1 = i-th position carries a i i I 1 = i-th position carries b i T I2 : i I 2 = (n i+1)-th position carries a i i I 2 = (n i+1)-th position carries c i Note: S I1 T I2 = I 1 I 2 = Restrict attention to input instances D(I, I) = (S I, T I ) for I {1,.., n}. (particular yes -instances) NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 27/55
73 A lower bound proof for DISJ n (1/5) DISJ n D n := {a 1, b 1, c 1..., a n, b n, c n} Input: two streams S = s 1, s 2,..., s n, T = t 1, t 2,..., t n of elements in D n, such that s i {a i, b i } and t n i+1 {a i, c i }. Question: Is {s 1, s 2,..., s n} {t 1, t 2,..., t n} =? Theorem: (Bar Yossef, Shalem, ICDE 08) DISJ n cannot be solved by a deterministic algorithm that performs one pass over each stream and that uses less than n log n 1 bits of memory. Proof: Consider input instances D(I 1, I 2 ) := (S I1, T I2 ) with I 1, I 2 {1,.., n} and S I1 : i I 1 = i-th position carries a i i I 1 = i-th position carries b i T I2 : i I 2 = (n i+1)-th position carries a i i I 2 = (n i+1)-th position carries c i Note: S I1 T I2 = I 1 I 2 = Restrict attention to input instances D(I, I) = (S I, T I ) for I {1,.., n}. (particular yes -instances) NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 27/55
74 A lower bound proof for DISJ n (2/5) Situation during a computation: input S: i n 1 n input T: n i+1 n 1 n memory buffer potential head positions: (i, j) with 1 i, j n start: (1, 1) n end: (n, n) T n For each input D(I, I) there exists exactly one i {1,.., n} such that the heads visit position (i, n i+1). NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 28/55 S
75 A lower bound proof for DISJ n (3/5) Goal now: cut-and-paste argument Find I, J {1,.., n} such that computations on D(I, I) and D(J, J) can be combined to an accepting computation on D(I, J ) for I and J with I J. = accept a no -instance! (1) Ex. i {1,.., n} and X 1 {I : I {1,.., n}} such that for each I X 1, head position (i, n i+1) is visited, X 1 2n. n (2) Ex. X 2 X 1 such that for all I, J X 2 : i I i J, X 2 X 1 2 2n 2n. (3) Ex. memory configuration c and X 3 X 2 such that for all I X 3 : memory configuration c when at head position (i, n i+1), X 3 X 2 m 2n 2nm. Note: X 3 > 1 m < 2n 2n s = log m < n log n 1. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 29/55
76 A lower bound proof for DISJ n (3/5) Goal now: cut-and-paste argument Find I, J {1,.., n} such that computations on D(I, I) and D(J, J) can be combined to an accepting computation on D(I, J ) for I and J with I J. = accept a no -instance! (1) Ex. i {1,.., n} and X 1 {I : I {1,.., n}} such that for each I X 1, head position (i, n i+1) is visited, X 1 2n. n (2) Ex. X 2 X 1 such that for all I, J X 2 : i I i J, X 2 X 1 2 2n 2n. (3) Ex. memory configuration c and X 3 X 2 such that for all I X 3 : memory configuration c when at head position (i, n i+1), X 3 X 2 m 2n 2nm. Note: X 3 > 1 m < 2n 2n s = log m < n log n 1. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 29/55
77 A lower bound proof for DISJ n (3/5) Goal now: cut-and-paste argument Find I, J {1,.., n} such that computations on D(I, I) and D(J, J) can be combined to an accepting computation on D(I, J ) for I and J with I J. = accept a no -instance! (1) Ex. i {1,.., n} and X 1 {I : I {1,.., n}} such that for each I X 1, head position (i, n i+1) is visited, X 1 2n. n (2) Ex. X 2 X 1 such that for all I, J X 2 : i I i J, X 2 X 1 2 2n 2n. (3) Ex. memory configuration c and X 3 X 2 such that for all I X 3 : memory configuration c when at head position (i, n i+1), X 3 X 2 m 2n 2nm. Note: X 3 > 1 m < 2n 2n s = log m < n log n 1. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 29/55
78 A lower bound proof for DISJ n (3/5) Goal now: cut-and-paste argument Find I, J {1,.., n} such that computations on D(I, I) and D(J, J) can be combined to an accepting computation on D(I, J ) for I and J with I J. = accept a no -instance! (1) Ex. i {1,.., n} and X 1 {I : I {1,.., n}} such that for each I X 1, head position (i, n i+1) is visited, X 1 2n. n (2) Ex. X 2 X 1 such that for all I, J X 2 : i I i J, X 2 X 1 2 2n 2n. (3) Ex. memory configuration c and X 3 X 2 such that for all I X 3 : memory configuration c when at head position (i, n i+1), X 3 X 2 m 2n 2nm. Note: X 3 > 1 m < 2n 2n s = log m < n log n 1. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 29/55
79 A lower bound proof for DISJ n (3/5) Goal now: cut-and-paste argument Find I, J {1,.., n} such that computations on D(I, I) and D(J, J) can be combined to an accepting computation on D(I, J ) for I and J with I J. = accept a no -instance! (1) Ex. i {1,.., n} and X 1 {I : I {1,.., n}} such that for each I X 1, head position (i, n i+1) is visited, X 1 2n. n (2) Ex. X 2 X 1 such that for all I, J X 2 : i I i J, X 2 X 1 2 2n 2n. (3) Ex. memory configuration c and X 3 X 2 such that for all I X 3 : memory configuration c when at head position (i, n i+1), X 3 X 2 m 2n 2nm. Note: X 3 > 1 m < 2n 2n s = log m < n log n 1. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 29/55
80 A lower bound proof for DISJ n (3/5) Goal now: cut-and-paste argument Find I, J {1,.., n} such that computations on D(I, I) and D(J, J) can be combined to an accepting computation on D(I, J ) for I and J with I J. = accept a no -instance! (1) Ex. i {1,.., n} and X 1 {I : I {1,.., n}} such that for each I X 1, head position (i, n i+1) is visited, X 1 2n. n (2) Ex. X 2 X 1 such that for all I, J X 2 : i I i J, X 2 X 1 2 2n 2n. (3) Ex. memory configuration c and X 3 X 2 such that for all I X 3 : memory configuration c when at head position (i, n i+1), X 3 X 2 m 2n 2nm. Note: X 3 > 1 = m < 2n 2n s = log m < n log n 1. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 29/55
81 A lower bound proof for DISJ n (4/5) Let I, J X 3 with I J. Same situation on input D(I, I) and on input D(J, J): input S: i n 1 n input T: n i+1 n 1 n memory buffer same memory configuration c Cut-and-paste argument = Same situation on inputs D(I 1, I 2 ) and D(I 1, I 2) I 1 = ( I {1,.., i 1} ) ( I {i} ) ( J {i+1,.., n} ) I 2 = ( I {i+1,.., n} ) ( I {i} ) ( J {1,.., i 1} ) I 1 = ( J {1,.., i 1} ) ( J {i} ) ( I {i+1,.., n} ) I 2 = ( J {i+1,.., n} ) ( J {i} ) ( I {1,.., i 1} ) Since I J, D(I 1, I 2 ) or D(I 1, I 2) is a no -instance. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 30/55
82 A lower bound proof for DISJ n (4/5) Let I, J X 3 with I J. Same situation on input D(I, I) and on input D(J, J): input S: i n 1 n input T: n i+1 n 1 n memory buffer same memory configuration c Cut-and-paste argument = Same situation on inputs D(I 1, I 2 ) and D(I 1, I 2) I 1 = ( I {1,.., i 1} ) ( I {i} ) ( J {i+1,.., n} ) I 2 = ( I {i+1,.., n} ) ( I {i} ) ( J {1,.., i 1} ) I 1 = ( J {1,.., i 1} ) ( J {i} ) ( I {i+1,.., n} ) I 2 = ( J {i+1,.., n} ) ( J {i} ) ( I {1,.., i 1} ) Since I J, D(I 1, I 2 ) or D(I 1, I 2) is a no -instance. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 30/55
83 A lower bound proof for DISJ n (4/5) Let I, J X 3 with I J. Same situation on input D(I, I) and on input D(J, J): input S: i n 1 n input T: n i+1 n 1 n memory buffer same memory configuration c Cut-and-paste argument = Same situation on inputs D(I 1, I 2 ) and D(I 1, I 2) I 1 = ( I {1,.., i 1} ) ( I {i} ) ( J {i+1,.., n} ) I 2 = ( I {i+1,.., n} ) ( I {i} ) ( J {1,.., i 1} ) I 1 = ( J {1,.., i 1} ) ( J {i} ) ( I {i+1,.., n} ) I 2 = ( J {i+1,.., n} ) ( J {i} ) ( I {1,.., i 1} ) Since I J, D(I 1, I 2 ) or D(I 1, I 2) is a no -instance. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 30/55
84 A lower bound proof for DISJ n (5/5) We have proved Theorem: (Bar Yossef, Shalem, ICDE 08) DISJ n cannot be solved by a deterministic algorithm that performs one pass over each stream and that uses less than n log n 1 bits of memory. The proof given by Bar-Yossef and Shalem (ICDE 2008) is different. For their proof, they introduce a particular kind of communication model: the token-based mesh communication model. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 31/55
85 Several passes over several streams in parallel General scenario: mp2s-automaton A with parameters (D, m, k f, k b ) input S: input T: memory buffer Parameters: 2 input streams: S = s 1, s 2,..., s n and T = t 1, t 2,..., t n of elements in D. m : number of possible memory configurations; s := log m size of the memory buffer (number of bits). k f forward heads on each input stream, k b backward heads on each input stream Depending on (a) the current memory state and (b) the elements in S and T at the current head positions, a deterministic transition function determines (1) the next memory state and (2) which of the heads should be advanced to the next position. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 32/55
86 Several passes over several streams in parallel General scenario: mp2s-automaton A with parameters (D, m, k f, k b ) input S: input T: memory buffer Parameters: 2 input streams: S = s 1, s 2,..., s n and T = t 1, t 2,..., t n of elements in D. m : number of possible memory configurations; s := log m size of the memory buffer (number of bits). k f forward heads on each input stream, k b backward heads on each input stream Depending on (a) the current memory state and (b) the elements in S and T at the current head positions, a deterministic transition function determines (1) the next memory state and (2) which of the heads should be advanced to the next position. NICOLE SCHWEIKARDT LOWER BOUNDS FOR MULTI-PASS PROCESSING OF MULTIPLE DATA STREAMS 32/55
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