Multi-join Query Evaluation on Big Data Lecture 2

Transcription

1 Multi-join Query Evaluation on Big Data Lecture 2 Dan Suciu March, 2015 Dan Suciu Multi-Joins Lecture 2 March, / 34

2 Multi-join Query Evaluation Outline Part 1 Optimal Sequential Algorithms. Thursday 14:15-15:45 Part 2 Lower bounds for Parallel Algorithms. Friday 14:15-15:45 Part 3 Optimal Parallel Algorithms. Saturday 9-10:30 Part 3 Data Skew. Saturday 11-12:30 Dan Suciu Multi-Joins Lecture 2 March, / 34

3 Brief Review of the AGM Bound Q(x) = R 1 (x 1 ),..., R l (x l ) Equal Cardinalities: R 1 =... R l = m Q m ρ, where ρ = fractional edge covering number of Q. General Case: R 1 = m 1,..., R l = m l Q min u m u 1 1 mu l l Dan Suciu Multi-Joins Lecture 2 March, / 34

4 Outline for Lecture 2 Background: Parallel Databases The MPC Model Algorithm for Triangles Lower Bound on the Communication Cost Lower Bound for General Queries Summary Dan Suciu Multi-Joins Lecture 2 March, / 34

5 Parallel Query Processing: Overview Parallel Databases: Queries and updates GAMMA machine (80 s), today in all commercial DBMS. Typically x10 s nodes. FO in AC 0 ; SQL embarrassingly parallel Parallel Data Analytics: Queries only this course MapReduce, Hadoop, PigLatin, Dremel, Scope, Spark. Typically x100 s or x1000 s nodes. Data reshuffling / communication is new bottleneck. Distributed State: Updates (transactions) will not discuss Replicate objects (3-5 times). Central problem: consistency. E.g. Google, Amazon, Yahoo, Microsoft, Ebay. Eventual consistency (NoSQL, Dynamo, BigTable, Peanuts). Strong consistency: Paxos, Spanner. Dan Suciu Multi-Joins Lecture 2 March, / 34

6 Parallel Query Processing: Overview Parallel Databases: Queries and updates GAMMA machine (80 s), today in all commercial DBMS. Typically x10 s nodes. FO in AC 0 ; SQL embarrassingly parallel Parallel Data Analytics: Queries only this course MapReduce, Hadoop, PigLatin, Dremel, Scope, Spark. Typically x100 s or x1000 s nodes. Data reshuffling / communication is new bottleneck. Distributed State: Updates (transactions) will not discuss Replicate objects (3-5 times). Central problem: consistency. E.g. Google, Amazon, Yahoo, Microsoft, Ebay. Eventual consistency (NoSQL, Dynamo, BigTable, Peanuts). Strong consistency: Paxos, Spanner. Dan Suciu Multi-Joins Lecture 2 March, / 34

7 Parallel Query Processing: Overview Parallel Databases: Queries and updates GAMMA machine (80 s), today in all commercial DBMS. Typically x10 s nodes. FO in AC 0 ; SQL embarrassingly parallel Parallel Data Analytics: Queries only this course MapReduce, Hadoop, PigLatin, Dremel, Scope, Spark. Typically x100 s or x1000 s nodes. Data reshuffling / communication is new bottleneck. Distributed State: Updates (transactions) will not discuss Replicate objects (3-5 times). Central problem: consistency. E.g. Google, Amazon, Yahoo, Microsoft, Ebay. Eventual consistency (NoSQL, Dynamo, BigTable, Peanuts). Strong consistency: Paxos, Spanner. Dan Suciu Multi-Joins Lecture 2 March, / 34

8 Key Concepts Input data of size m is partitioned on p servers, connected by a network. Query processing involves local computation and communication. Performance parameters [Gray&Dewitt 92] Speedup How does performance change when p increases? Ideal: linear speedup. Scaleup How does performance chance when p, m increase at same rate? Ideal: constant scaleup. Speed Speed # processors (=P) # processors (=P) AND data size Dan Suciu Multi-Joins Lecture 2 March, / 34

9 Data Partition Given a database of size m, partition it on p servers. Balanced partition Each server holds m/p data. Skewed partition Some server holds m/p data. Usually, the input data is already partitioned, but we need to re-partition for a particular problem. Data reshuffling. Dan Suciu Multi-Joins Lecture 2 March, / 34

10 Example 1: Hash-Partitioned Join Compute Q(x, y, z) = R(x, y) S(y, z), where R = m 1, S = m 2 Input data The input relations R, S are partitioned on the servers. Data reshuffling Given hash function h Domain [p]: Send every tuple R(x, y) to server h(y). Send every tuple S(y, z) to server h(y). Computation In parallel, each server i computes the join R i (x, y) S i (y, z) of its local fragments R i, S i. If y is a key, then the load/server is m 1 /p + m 2 /p (why?) Linear Speedup Otherwise, we may have skew (why?) What is the worst speedup? Dan Suciu Multi-Joins Lecture 2 March, / 34

11 Understanding Skew Given: R(x, y) with m items, p servers. Send each tuple R(x, y) to server h(y), where h = random function. Claim 1 For each server u, its expected load is E[L u ] = m/p. Why? Claim 2 We say that R is skewed if some value y occurs more than m/p times in R. Then some server has L u > m/p. Claim 3 If R is not skewed, the max u L u is O(m/p) with high probability. (Details in lecture 4.) Take-away: we assume no skew, then: Max-load = O(Expected load) Dan Suciu Multi-Joins Lecture 2 March, / 34

12 Understanding Skew Given: R(x, y) with m items, p servers. Send each tuple R(x, y) to server h(y), where h = random function. Claim 1 For each server u, its expected load is E[L u ] = m/p. Why? Claim 2 We say that R is skewed if some value y occurs more than m/p times in R. Then some server has L u > m/p. Claim 3 If R is not skewed, the max u L u is O(m/p) with high probability. (Details in lecture 4.) Take-away: we assume no skew, then: Max-load = O(Expected load) Dan Suciu Multi-Joins Lecture 2 March, / 34

13 Understanding Skew Given: R(x, y) with m items, p servers. Send each tuple R(x, y) to server h(y), where h = random function. Claim 1 For each server u, its expected load is E[L u ] = m/p. Why? Claim 2 We say that R is skewed if some value y occurs more than m/p times in R. Then some server has L u > m/p. Claim 3 If R is not skewed, the max u L u is O(m/p) with high probability. (Details in lecture 4.) Take-away: we assume no skew, then: Max-load = O(Expected load) Dan Suciu Multi-Joins Lecture 2 March, / 34

14 Understanding Skew Given: R(x, y) with m items, p servers. Send each tuple R(x, y) to server h(y), where h = random function. Claim 1 For each server u, its expected load is E[L u ] = m/p. Why? Claim 2 We say that R is skewed if some value y occurs more than m/p times in R. Then some server has L u > m/p. Claim 3 If R is not skewed, the max u L u is O(m/p) with high probability. (Details in lecture 4.) Take-away: we assume no skew, then: Max-load = O(Expected load) Dan Suciu Multi-Joins Lecture 2 March, / 34

15 Example 2: Broadcast Join Compute Q(x, y, z) = R(x, y) S(y, z), where R = m 1 S = m 2 Input data The input relations R, S are partitioned on the servers. Broadcast Send every tuple S(y, z) to every server Computation In parallel, each server u computes the join R u (x, y) S(y, z) of its local fragment R u with S. If m 2 m 1 /p then the Broadcast Join is very effective. Used a lot in practice. Dan Suciu Multi-Joins Lecture 2 March, / 34

16 Massively Parallel Communication Model (MPC) [Beame 13] The MPC model is the following: p servers are connected by a network. Servers are infinitely powerful. Input Data of size m is initially uniformly partitioned. Computation = several rounds. One round = local computation plus global communication. Dan Suciu Multi-Joins Lecture 2 March, / 34

17 Massively Parallel Communication Model (MPC) [Beame 13] The MPC model is the following: p servers are connected by a network. Servers are infinitely powerful. Input Data of size m is initially uniformly partitioned. Computation = several rounds. One round = local computation plus global communication. The only cost is the communication. Definition (The Load of a algorithm on the MPC Model) L u = maximum amount of data received by server u during any round L = max u L u Dan Suciu Multi-Joins Lecture 2 March, / 34

18 Massively Parallel Communication Model (MPC) Number of servers = p Input (size=m) O(m/p) O(m/p) Server Server p Input data = size m Algorithm = Several rounds L Server 1 L.... L Round 1 Server p L Round 2 One round = Compute & communicate Server 1 Max communication load per server = L L Server p L Round 3 Dan Suciu Multi-Joins Lecture 2 March, / 34

19 Load/Rounds Tradeoff in the MPC Model Naive 1-Round Send entire data to server 1, compute locally. L = m Naive p-rounds At each round, send a m/p-fragment of the data to server 1, then compute locally. L = m/p. Ideal Algorithms 1-Round, load L = m/p (but rarely possible) Real Algorithms O(1) rounds, and L = O(m/p 1 ε ), for 0 ε < 1. Dan Suciu Multi-Joins Lecture 2 March, / 34

20 Load/Rounds Tradeoff in the MPC Model Naive 1-Round Send entire data to server 1, compute locally. L = m Naive p-rounds At each round, send a m/p-fragment of the data to server 1, then compute locally. L = m/p. Ideal Algorithms 1-Round, load L = m/p (but rarely possible) Real Algorithms O(1) rounds, and L = O(m/p 1 ε ), for 0 ε < 1. Dan Suciu Multi-Joins Lecture 2 March, / 34

21 Load/Rounds Tradeoff in the MPC Model Naive 1-Round Send entire data to server 1, compute locally. L = m Naive p-rounds At each round, send a m/p-fragment of the data to server 1, then compute locally. L = m/p. Ideal Algorithms 1-Round, load L = m/p (but rarely possible) Real Algorithms O(1) rounds, and L = O(m/p 1 ε ), for 0 ε < 1. Dan Suciu Multi-Joins Lecture 2 March, / 34

22 Load/Rounds Tradeoff in the MPC Model Naive 1-Round Send entire data to server 1, compute locally. L = m Naive p-rounds At each round, send a m/p-fragment of the data to server 1, then compute locally. L = m/p. Ideal Algorithms 1-Round, load L = m/p (but rarely possible) Real Algorithms O(1) rounds, and L = O(m/p 1 ε ), for 0 ε < 1. Dan Suciu Multi-Joins Lecture 2 March, / 34

23 Examples on the MPC Model Example: Join Q(x, y, z) = R(x, y), S(y, z). Round 1 (Hash-partitioned join) Send tuple R(x, y) to server h(y), send S(y, z) to server h(y), compute the join locally. Load: O(m/p) (assuming no skew). Example: Triangles Q(x, y, z) = R(x, y), S(y, z), T (z, x) Round 1 hash-partitioned join: Aux(x, y, z) = R(x, y), S(y, z) Round 2 hash-partitioned join: Q(x, y, z) = Aux(x, y, z), T (z, x) Load: can be as high as m 2 /p because of the intermediate result!! Can we compute triangles with a smaller load? Dan Suciu Multi-Joins Lecture 2 March, / 34

24 Examples on the MPC Model Example: Join Q(x, y, z) = R(x, y), S(y, z). Round 1 (Hash-partitioned join) Send tuple R(x, y) to server h(y), send S(y, z) to server h(y), compute the join locally. Load: O(m/p) (assuming no skew). Example: Triangles Q(x, y, z) = R(x, y), S(y, z), T (z, x) Round 1 hash-partitioned join: Aux(x, y, z) = R(x, y), S(y, z) Round 2 hash-partitioned join: Q(x, y, z) = Aux(x, y, z), T (z, x) Load: can be as high as m 2 /p because of the intermediate result!! Can we compute triangles with a smaller load? Dan Suciu Multi-Joins Lecture 2 March, / 34

25 A Simple Lower Bound for the MPC Model Let Q be a query, with R 1 = = R l = m. Fact For any algorithm for Q with r rounds and load L, it holds: r L m/p 1/ρ. Dan Suciu Multi-Joins Lecture 2 March, / 34

26 A Simple Lower Bound for the MPC Model Let Q be a query, with R 1 = = R l = m. Fact For any algorithm for Q with r rounds and load L, it holds: r L m/p 1/ρ. Proof Construct an instance where Q = m ρ. One server: receives r L tuples from R j, for all j, hence finds (r L) ρ answers. All p servers find p(r L) ρ answers. It follows: r L m/p 1/ρ. Dan Suciu Multi-Joins Lecture 2 March, / 34

27 A Simple Lower Bound for the MPC Model Let Q be a query, with R 1 = = R l = m. Fact For any algorithm for Q with r rounds and load L, it holds: r L m/p 1/ρ. Proof Construct an instance where Q = m ρ. One server: receives r L tuples from R j, for all j, hence finds (r L) ρ answers. All p servers find p(r L) ρ answers. It follows: r L m/p 1/ρ. Question For Q = R(x, y), S(y, z) it should be r L m/p 1/2, but we computed a join with load L = m/p. Contradiction? Dan Suciu Multi-Joins Lecture 2 March, / 34

28 A Simple Lower Bound for the MPC Model Let Q be a query, with R 1 = = R l = m. Fact For any algorithm for Q with r rounds and load L, it holds: r L m/p 1/ρ. Proof Construct an instance where Q = m ρ. One server: receives r L tuples from R j, for all j, hence finds (r L) ρ answers. All p servers find p(r L) ρ answers. It follows: r L m/p 1/ρ. Question For Q = R(x, y), S(y, z) it should be r L m/p 1/2, but we computed a join with load L = m/p. Contradiction? Answer: the algorithm is for skew-free data, the bound for skewed data. Dan Suciu Multi-Joins Lecture 2 March, / 34

29 One-Round Algorithm for Triangles [Afrati&Ullman 10,Suri&Vassilvitski 11] Q(x, y, z) = R(x, y), S(y, z), T (z, x) Place the p servers in a cube: [p] [p 1/3 ] [p 1/3 ] [p 1/3 ]. Round 1 In parallel, each server does the following: Send R(x, y) to all servers (h 1 (x), h 2 (y), ) Send S(y, z) to all servers (, h 2 (y), h 3 (z)) Send T (z, x) to all servers (h 1 (x),, h 3 (z)) Then compute Q locally Theorem (Beame 14) (1) If h 1, h 2, h 3 are independent hash functions, then the expected load at some server u is E[L u ] = m 1+m 2 +m 3 def = L. [Why do we need independence?] p 2/3 (2) If the data has no skew, then maximum load is O(L) w.h.p. Dan Suciu Multi-Joins Lecture 2 March, / 34

30 One-Round Algorithm for Triangles [Afrati&Ullman 10,Suri&Vassilvitski 11] Q(x, y, z) = R(x, y), S(y, z), T (z, x) Place the p servers in a cube: [p] [p 1/3 ] [p 1/3 ] [p 1/3 ]. Round 1 In parallel, each server does the following: Send R(x, y) to all servers (h 1 (x), h 2 (y), ) Send S(y, z) to all servers (, h 2 (y), h 3 (z)) Send T (z, x) to all servers (h 1 (x),, h 3 (z)) Then compute Q locally Theorem (Beame 14) (1) If h 1, h 2, h 3 are independent hash functions, then the expected load at some server u is E[L u ] = m 1+m 2 +m 3 def = L. [Why do we need independence?] p 2/3 (2) If the data has no skew, then maximum load is O(L) w.h.p. Dan Suciu Multi-Joins Lecture 2 March, / 34

31 One-Round Algorithm for Triangles [Afrati&Ullman 10,Suri&Vassilvitski 11] Q(x, y, z) = R(x, y), S(y, z), T (z, x) Place the p servers in a cube: [p] [p 1/3 ] [p 1/3 ] [p 1/3 ]. Round 1 In parallel, each server does the following: Send R(x, y) to all servers (h 1 (x), h 2 (y), ) Send S(y, z) to all servers (, h 2 (y), h 3 (z)) Send T (z, x) to all servers (h 1 (x),, h 3 (z)) Then compute Q locally Theorem (Beame 14) (1) If h 1, h 2, h 3 are independent hash functions, then the expected load at some server u is E[L u ] = m 1+m 2 +m 3 def = L. [Why do we need independence?] p 2/3 (2) If the data has no skew, then maximum load is O(L) w.h.p. Dan Suciu Multi-Joins Lecture 2 March, / 34

32 Discussion We will call the algorithm HyperCube, following [Beame 13]. Each tuple R(x, y) is replicated p 1/3 times. Hence, load L = m/p 2/3 Partitioning R(x, y) (h 1 (x), h 2 (y), ) more tolerant to skew: Each value x is sent to p 1/3 buckets, each y is sent to p 1/3 buckets. Can tolerate degrees up to m/p 1/3 (better than m/p). Notice that the algorithm only shuffles the data: each server still has to compute the query locally. Important to use worst-case algorithm. Non-linear speedup, because the load is m/p 2/3. Can we we compute triangles with a load of m/p? Dan Suciu Multi-Joins Lecture 2 March, / 34

33 Lower Bound for Triangle Queries Q(x, y, z) = R(x, y), S(y, z), T (z, x) Assume R + S + T = m. Denote L lower = m/p 2/3. Theorem Any 1-round algorithm that computes Q must have load L L lower, even on database instances without skew. Hence, HyperCube is optimal for computing triangles on permutations. Next, we will prove the theorem. Dan Suciu Multi-Joins Lecture 2 March, / 34

34 Lower Bound for Triangle Queries Q(x, y, z) = R(x, y), S(y, z), T (z, x) We assume R, S, T are permutations over a domain of size n. Example n = 4: R x y S y z T z x = Q x y z Question: what is E[ Q ], over random permutations R, S, T? Theorem (Beame 13) Denote L lower = 3n/p 2/3. Let A be an algorithm for Q, with load L < L lower. Then, the expected number of triangles returned by A is E[ A ] ( L ) L lower 3/2 E[ Q ] Dan Suciu Multi-Joins Lecture 2 March, / 34

35 Lower Bound for Triangle Queries Q(x, y, z) = R(x, y), S(y, z), T (z, x) We assume R, S, T are permutations over a domain of size n. Example n = 4: R x y S y z T z x = Q x y z Question: what is E[ Q ], over random permutations R, S, T? Theorem (Beame 13) Denote L lower = 3n/p 2/3. Let A be an algorithm for Q, with load L < L lower. Then, the expected number of triangles returned by A is E[ A ] ( L ) L lower 3/2 E[ Q ] Dan Suciu Multi-Joins Lecture 2 March, / 34

36 Lower Bound for Triangle Queries Q(x, y, z) = R(x, y), S(y, z), T (z, x) We assume R, S, T are permutations over a domain of size n. Example n = 4: R x y S y z T z x = Q x y z Question: what is E[ Q ], over random permutations R, S, T? Theorem (Beame 13) Denote L lower = 3n/p 2/3. Let A be an algorithm for Q, with load L < L lower. Then, the expected number of triangles returned by A is E[ A ] ( L ) L lower 3/2 E[ Q ] Dan Suciu Multi-Joins Lecture 2 March, / 34

37 Discussion: Inputs, Messages, Bits Initially, the relations R, S, T are on disjoint servers. This is w.l.o.g. Stronger: assume that R is on server 1, S on server 2, T on server 3. Any server u receives three messages: msg 1 about R msg 2 about S msg 3 about T msg 1 + msg 2 + msg 3 L bits. Messages may encode arbitrary information. E.g. bit 1 says R is even ; bit 2 says R has a 17-cycle ; etc. No lower bound is possible for a fixed input R, S, T : an algorithm may simply check for that input and encode using L = O(1) bits. Instead, the lower bound is a statement about all permutations. Dan Suciu Multi-Joins Lecture 2 March, / 34

38 Notation: Bits v.s. Tuples Size of db in #tuples m = R + S + T. Size of db in #bits M = R log n + S log n + T log n = m log n where n = size of the domain. If R is a random permutation, then size(r) = log(n!) n log n bits. L lower = m/p 2/3 tuples becomes L lower = 3 log(n!)/p 2/3 bits. Will prove the following, where L, L lower are expressed in bits: L E[ A ] ( L lower ) 3/2 E[ Q ] Dan Suciu Multi-Joins Lecture 2 March, / 34

39 Proof Part 1: E[ Q ] on Random Inputs Q(x, y, z) = R(x, y), S(y, z), T (z, x) Lemma E[ Q ] = 1, where the expectation is over random permutations R, S, T. Proof. Note: i, j, k [n], P((i, j) R) = P((j, k) S) = P((k, i) T ) = 1/n. E[ Q ] = P((i, j) R (j, k) S (k, i) T ) i,j,k independence = P((i, j) R) P((j, k) S) P((k, i) T ) = n 3 1 i,j,k n 1 n 1 n = 1 Dan Suciu Multi-Joins Lecture 2 March, / 34

40 Proof Part 1: E[ Q ] on Random Inputs Q(x, y, z) = R(x, y), S(y, z), T (z, x) Lemma E[ Q ] = 1, where the expectation is over random permutations R, S, T. Proof. Note: i, j, k [n], P((i, j) R) = P((j, k) S) = P((k, i) T ) = 1/n. E[ Q ] = P((i, j) R (j, k) S (k, i) T ) i,j,k independence = P((i, j) R) P((j, k) S) P((k, i) T ) = n 3 1 i,j,k n 1 n 1 n = 1 Dan Suciu Multi-Joins Lecture 2 March, / 34

41 Proof Part 1: E[ Q ] on Random Inputs Q(x, y, z) = R(x, y), S(y, z), T (z, x) Lemma E[ Q ] = 1, where the expectation is over random permutations R, S, T. Proof. Note: i, j, k [n], P((i, j) R) = P((j, k) S) = P((k, i) T ) = 1/n. E[ Q ] = P((i, j) R (j, k) S (k, i) T ) i,j,k independence = P((i, j) R) P((j, k) S) P((k, i) T ) = n 3 1 i,j,k n 1 n 1 n = 1 Dan Suciu Multi-Joins Lecture 2 March, / 34

42 Discussion In expectation, there is one triangle! E[ Q ] = 1 Will prove: if algorithm A has load L < L lower, then E[A] = (L/L lower ) 3/2. Dan Suciu Multi-Joins Lecture 2 March, / 34

43 Proof Part 2: What We Learn from a Message Fix a server u [p], and a message msg 1 it received about R. Definition The set of known tuples is: K 1 (msg 1 ) = {(i, j) R(msg 1 (R) = msg 1 (i, j) R)} Similarly K 2 (msg 2 ), K 3 (msg 3 ) known tuples in S and T. Observation Upon receiving msg 1, msg 2, msg 3, server u can output triangle (i, j, k) iff (i, j) K 1 (msg 1 ) (j, k) K 2 (msg 2 ) (k, i) K 2 (msg 3 ) Dan Suciu Multi-Joins Lecture 2 March, / 34

44 Proof Part 2: What We Learn from a Message Fix a server u [p], and a message msg 1 it received about R. Definition The set of known tuples is: K 1 (msg 1 ) = {(i, j) R(msg 1 (R) = msg 1 (i, j) R)} Similarly K 2 (msg 2 ), K 3 (msg 3 ) known tuples in S and T. Observation Upon receiving msg 1, msg 2, msg 3, server u can output triangle (i, j, k) iff (i, j) K 1 (msg 1 ) (j, k) K 2 (msg 2 ) (k, i) K 2 (msg 3 ) Dan Suciu Multi-Joins Lecture 2 March, / 34

45 Proof Part 2: What We Learn from a Message Fix a server u [p], and a message msg 1 it received about R. Definition The set of known tuples is: K 1 (msg 1 ) = {(i, j) R(msg 1 (R) = msg 1 (i, j) R)} Similarly K 2 (msg 2 ), K 3 (msg 3 ) known tuples in S and T. Observation Upon receiving msg 1, msg 2, msg 3, server u can output triangle (i, j, k) iff (i, j) K 1 (msg 1 ) (j, k) K 2 (msg 2 ) (k, i) K 2 (msg 3 ) Dan Suciu Multi-Joins Lecture 2 March, / 34

46 Discussion The useful information we learn from a message is the set of known tuples. We show next: if L 1 = size(msg 1 ) is small, then K 1 (msg 1 ) is small If you know only a few bits, then you know only a few tuples Later: if K 1, K 2, K 3 are small, the server knows only few triangles (AGM) Dan Suciu Multi-Joins Lecture 2 March, / 34

47 Proof Part 3: With Few Bits You Know Only Few Tuples Denote H(R) = log(n!) the entropy of R. Proposition Let f 1 < 1. If msg 1 (R) has f 1 H(R) bits, then E[ K 1 (msg 1 (R)) ] f 1 n, where the expectation is over random permutations R. Dan Suciu Multi-Joins Lecture 2 March, / 34

48 Proof Part 3: With Few Bits You Know Only Few Tuples Denote H(R) = log(n!) the entropy of R. Proposition Let f 1 < 1. If msg 1 (R) has f 1 H(R) bits, then E[ K 1 (msg 1 (R)) ] f 1 n, where the expectation is over random permutations R. Proof. Denote k = K 1(m 1). Dan Suciu Multi-Joins Lecture 2 March, / 34

49 Proof Part 3: With Few Bits You Know Only Few Tuples Denote H(R) = log(n!) the entropy of R. Proposition Let f 1 < 1. If msg 1 (R) has f 1 H(R) bits, then E[ K 1 (msg 1 (R)) ] f 1 n, where the expectation is over random permutations R. Proof. Denote k = K 1(m 1). H(R m 1) log[(n k)!] (1 k ) H(R) because log[(n k)!]/(n k) log(n!)/n. n Dan Suciu Multi-Joins Lecture 2 March, / 34

50 Proof Part 3: With Few Bits You Know Only Few Tuples Denote H(R) = log(n!) the entropy of R. Proposition Let f 1 < 1. If msg 1 (R) has f 1 H(R) bits, then E[ K 1 (msg 1 (R)) ] f 1 n, where the expectation is over random permutations R. Proof. Denote k = K 1(m 1). H(R m 1) log[(n k)!] (1 k ) H(R) because log[(n k)!]/(n k) log(n!)/n. n H(R) = H(R, msg 1 (R)) R determines msg 1 (R) Dan Suciu Multi-Joins Lecture 2 March, / 34

51 Proof Part 3: With Few Bits You Know Only Few Tuples Denote H(R) = log(n!) the entropy of R. Proposition Let f 1 < 1. If msg 1 (R) has f 1 H(R) bits, then E[ K 1 (msg 1 (R)) ] f 1 n, where the expectation is over random permutations R. Proof. Denote k = K 1(m 1). H(R m 1) log[(n k)!] (1 k ) H(R) because log[(n k)!]/(n k) log(n!)/n. n H(R) = H(R, msg 1 (R)) R determines msg 1 (R) = H(msg 1 (R)) + m1 H(R m 1) P(m 1) chain rule Dan Suciu Multi-Joins Lecture 2 March, / 34

52 Proof Part 3: With Few Bits You Know Only Few Tuples Denote H(R) = log(n!) the entropy of R. Proposition Let f 1 < 1. If msg 1 (R) has f 1 H(R) bits, then E[ K 1 (msg 1 (R)) ] f 1 n, where the expectation is over random permutations R. Proof. Denote k = K 1(m 1). H(R m 1) log[(n k)!] (1 k ) H(R) because log[(n k)!]/(n k) log(n!)/n. n H(R) = H(R, msg 1 (R)) R determines msg 1 (R) = H(msg 1 (R)) + m1 H(R m 1) P(m 1) chain rule f 1 H(R) + m1 (1 K 1(m 1 ) n ) H(R) P(m 1) K = f 1 H(R) + [1 1 (m 1 ) P(m 1 ) m1 ] H(R) = f n 1 H(R) + [1 E[ K 1(m 1 ) ] ] H(R) n Dan Suciu Multi-Joins Lecture 2 March, / 34

53 Proof Part 3: With Few Bits You Know Only Few Tuples Denote H(R) = log(n!) the entropy of R. Proposition Let f 1 < 1. If msg 1 (R) has f 1 H(R) bits, then E[ K 1 (msg 1 (R)) ] f 1 n, where the expectation is over random permutations R. Proof. Denote k = K 1(m 1). H(R m 1) log[(n k)!] (1 k ) H(R) because log[(n k)!]/(n k) log(n!)/n. n H(R) = H(R, msg 1 (R)) R determines msg 1 (R) = H(msg 1 (R)) + m1 H(R m 1) P(m 1) chain rule f 1 H(R) + m1 (1 K 1(m 1 ) n ) H(R) P(m 1) K = f 1 H(R) + [1 1 (m 1 ) P(m 1 ) m1 ] H(R) = f n 1 H(R) + [1 E[ K 1(m 1 ) ] ] H(R) n It follows: E[ K 1(m 1) ] f 1n Dan Suciu Multi-Joins Lecture 2 March, / 34

54 Proof Part 4: Few Known Tuples form Few Triangles Continue to fix one server u. Its load L = L 1 + L 2 + L 3. Denote: a ij = Pr((i, j) K 1 (msg 1 (R))); similarly b jk, c ki for S, T. Denote A u the set of triangles returned by u: E[ A u ] = i,j,k a ij b jk c ki (( ij a 2 ij )( jk b 2 jk )( ki c 2 ki ))1/2 (Friedgut) a ij 1/n (why?) and ij a ij = E[ K 1 (msg 1 (R)) ] It follows: ij a 2 ij 1/n ij a ij L 1 log(n!) L 1 log(n!) n (why?) E[ A u ] ( L 1 log(n!) L 2 log(n!) L 3 log(n!) )1/2 ( L 1+L 2 +L 3 3 log(n!) )3/2 = ( L M )3/2 triangles. 3/2 All p servers return E[ A ] p ( L M )3/2 = ( L M ) = ( L L lower ) 3/2 triangles. p 2/3 QED Dan Suciu Multi-Joins Lecture 2 March, / 34

55 Proof Part 4: Few Known Tuples form Few Triangles Continue to fix one server u. Its load L = L 1 + L 2 + L 3. Denote: a ij = Pr((i, j) K 1 (msg 1 (R))); similarly b jk, c ki for S, T. Denote A u the set of triangles returned by u: E[ A u ] = i,j,k a ij b jk c ki (( ij a 2 ij )( jk b 2 jk )( ki c 2 ki ))1/2 (Friedgut) a ij 1/n (why?) and ij a ij = E[ K 1 (msg 1 (R)) ] It follows: ij a 2 ij 1/n ij a ij L 1 log(n!) L 1 log(n!) n (why?) E[ A u ] ( L 1 log(n!) L 2 log(n!) L 3 log(n!) )1/2 ( L 1+L 2 +L 3 3 log(n!) )3/2 = ( L M )3/2 triangles. 3/2 All p servers return E[ A ] p ( L M )3/2 = ( L M ) = ( L L lower ) 3/2 triangles. p 2/3 QED Dan Suciu Multi-Joins Lecture 2 March, / 34

56 Proof Part 4: Few Known Tuples form Few Triangles Continue to fix one server u. Its load L = L 1 + L 2 + L 3. Denote: a ij = Pr((i, j) K 1 (msg 1 (R))); similarly b jk, c ki for S, T. Denote A u the set of triangles returned by u: E[ A u ] = i,j,k a ij b jk c ki (( ij a 2 ij )( jk b 2 jk )( ki c 2 ki ))1/2 (Friedgut) a ij 1/n (why?) and ij a ij = E[ K 1 (msg 1 (R)) ] It follows: ij a 2 ij 1/n ij a ij L 1 log(n!) L 1 log(n!) n (why?) E[ A u ] ( L 1 log(n!) L 2 log(n!) L 3 log(n!) )1/2 ( L 1+L 2 +L 3 3 log(n!) )3/2 = ( L M )3/2 triangles. 3/2 All p servers return E[ A ] p ( L M )3/2 = ( L M ) = ( L L lower ) 3/2 triangles. p 2/3 QED Dan Suciu Multi-Joins Lecture 2 March, / 34

57 Proof Part 4: Few Known Tuples form Few Triangles Continue to fix one server u. Its load L = L 1 + L 2 + L 3. Denote: a ij = Pr((i, j) K 1 (msg 1 (R))); similarly b jk, c ki for S, T. Denote A u the set of triangles returned by u: E[ A u ] = i,j,k a ij b jk c ki (( ij a 2 ij )( jk b 2 jk )( ki c 2 ki ))1/2 (Friedgut) a ij 1/n (why?) and ij a ij = E[ K 1 (msg 1 (R)) ] It follows: ij a 2 ij 1/n ij a ij L 1 log(n!) L 1 log(n!) n (why?) E[ A u ] ( L 1 log(n!) L 2 log(n!) L 3 log(n!) )1/2 ( L 1+L 2 +L 3 3 log(n!) )3/2 = ( L M )3/2 triangles. 3/2 All p servers return E[ A ] p ( L M )3/2 = ( L M ) = ( L L lower ) 3/2 triangles. p 2/3 QED Dan Suciu Multi-Joins Lecture 2 March, / 34

58 Proof Part 4: Few Known Tuples form Few Triangles Continue to fix one server u. Its load L = L 1 + L 2 + L 3. Denote: a ij = Pr((i, j) K 1 (msg 1 (R))); similarly b jk, c ki for S, T. Denote A u the set of triangles returned by u: E[ A u ] = i,j,k a ij b jk c ki (( ij a 2 ij )( jk b 2 jk )( ki c 2 ki ))1/2 (Friedgut) a ij 1/n (why?) and ij a ij = E[ K 1 (msg 1 (R)) ] It follows: ij a 2 ij 1/n ij a ij L 1 log(n!) L 1 log(n!) n (why?) E[ A u ] ( L 1 log(n!) L 2 log(n!) L 3 log(n!) )1/2 ( L 1+L 2 +L 3 3 log(n!) )3/2 = ( L M )3/2 triangles. 3/2 All p servers return E[ A ] p ( L M )3/2 = ( L M ) = ( L L lower ) 3/2 triangles. p 2/3 QED Dan Suciu Multi-Joins Lecture 2 March, / 34

59 Proof Part 4: Few Known Tuples form Few Triangles Continue to fix one server u. Its load L = L 1 + L 2 + L 3. Denote: a ij = Pr((i, j) K 1 (msg 1 (R))); similarly b jk, c ki for S, T. Denote A u the set of triangles returned by u: E[ A u ] = i,j,k a ij b jk c ki (( ij a 2 ij )( jk b 2 jk )( ki c 2 ki ))1/2 (Friedgut) a ij 1/n (why?) and ij a ij = E[ K 1 (msg 1 (R)) ] It follows: ij a 2 ij 1/n ij a ij L 1 log(n!) L 1 log(n!) n (why?) E[ A u ] ( L 1 log(n!) L 2 log(n!) L 3 log(n!) )1/2 ( L 1+L 2 +L 3 3 log(n!) )3/2 = ( L M )3/2 triangles. 3/2 All p servers return E[ A ] p ( L M )3/2 = ( L M ) = ( L L lower ) 3/2 triangles. p 2/3 QED Dan Suciu Multi-Joins Lecture 2 March, / 34

60 Proof Part 4: Few Known Tuples form Few Triangles Continue to fix one server u. Its load L = L 1 + L 2 + L 3. Denote: a ij = Pr((i, j) K 1 (msg 1 (R))); similarly b jk, c ki for S, T. Denote A u the set of triangles returned by u: E[ A u ] = i,j,k a ij b jk c ki (( ij a 2 ij )( jk b 2 jk )( ki c 2 ki ))1/2 (Friedgut) a ij 1/n (why?) and ij a ij = E[ K 1 (msg 1 (R)) ] It follows: ij a 2 ij 1/n ij a ij L 1 log(n!) L 1 log(n!) n (why?) E[ A u ] ( L 1 log(n!) L 2 log(n!) L 3 log(n!) )1/2 ( L 1+L 2 +L 3 3 log(n!) )3/2 = ( L M )3/2 triangles. 3/2 All p servers return E[ A ] p ( L M )3/2 = ( L M ) = ( L L lower ) 3/2 triangles. p 2/3 QED Dan Suciu Multi-Joins Lecture 2 March, / 34

61 Discussion of the Lower Bound for the Triangle Query Algorithm A with load L < L lower = m/p 2/3, E[A] (L/L lower ) 3/2 E[ Q ] The result is for Skew-free Data There exists at least one instance on which A fails (trivial). Even if A is randomized, there exists at least one instance where A fails with high probability (Yao s lemma). Parallelism gets harder as p increases. An algorithm with load L = O( m 1 ), with ε < 1/3 reports only O( ) triangles. Fewer, as p 1 ε p 1/3 ε p increases! Dan Suciu Multi-Joins Lecture 2 March, / 34

62 Discussion of the Lower Bound for the Triangle Query Algorithm A with load L < L lower = m/p 2/3, E[A] (L/L lower ) 3/2 E[ Q ] The result is for Skew-free Data There exists at least one instance on which A fails (trivial). Even if A is randomized, there exists at least one instance where A fails with high probability (Yao s lemma). Parallelism gets harder as p increases. An algorithm with load L = O( m 1 ), with ε < 1/3 reports only O( ) triangles. Fewer, as p 1 ε p 1/3 ε p increases! Dan Suciu Multi-Joins Lecture 2 March, / 34

63 Generalization to Full Conjunctive Queries Will discuss the equal-cardinality case today. Will discuss the general case in Lecture 3. Dan Suciu Multi-Joins Lecture 2 March, / 34

64 The Fractional Vertex Cover / Edge Packing Hypergraph: Q = R 1 (x 1 ),..., R l (x l ) Nodes: x 1,..., x k, edges: R 1,..., R l. Definition A fractional vertex cover of Q is a sequence v 1 0,..., v k 0 such that: j i x i R j v i 1 A fractional edge packing of Q is a sequence u 1 0,..., u l 0 such that: i j x i R j u j 1 By duality: min v i v i = max u j u j = τ Dan Suciu Multi-Joins Lecture 2 March, / 34

65 The Fractional Vertex Cover / Edge Packing Hypergraph: Q = R 1 (x 1 ),..., R l (x l ) Nodes: x 1,..., x k, edges: R 1,..., R l. Definition A fractional vertex cover of Q is a sequence v 1 0,..., v k 0 such that: j i x i R j v i 1 A fractional edge packing of Q is a sequence u 1 0,..., u l 0 such that: i j x i R j u j 1 By duality: min v i v i = max u j u j = τ Dan Suciu Multi-Joins Lecture 2 March, / 34

66 The HyperCube Algorithm for General Queries Q(x) = R 1 (x 1 ),..., R l (x l ), R 1 =... = R l = m p servers. Dan Suciu Multi-Joins Lecture 2 March, / 34

67 The HyperCube Algorithm for General Queries Q(x) = R 1 (x 1 ),..., R l (x l ), R 1 =... = R l = m p servers. def v = (v 1,..., v k ) any fractional vertex cover; v 0 = i v i. Organize the p servers in a hypercube: [p] [p v 1 v0 ] [p Choose k independent hash functions h 1,..., h k v k v0 ]. Dan Suciu Multi-Joins Lecture 2 March, / 34

68 The HyperCube Algorithm for General Queries Q(x) = R 1 (x 1 ),..., R l (x l ), R 1 =... = R l = m p servers. def v = (v 1,..., v k ) any fractional vertex cover; v 0 = i v i. Organize the p servers in a hypercube: [p] [p v 1 v k v0 ] [p v0 ]. Choose k independent hash functions h 1,..., h k Round 1 Each server sends each tuple R j (x j1, x j2,...) to all servers whose coordinates j 1, j 2,... are h j1 (x j1 ), h j2 (x j2 ),... and broadcasts along the missing dimensions. Then, each server computes Q on its local data. Dan Suciu Multi-Joins Lecture 2 March, / 34

69 The HyperCube Algorithm for General Queries Q(x) = R 1 (x 1 ),..., R l (x l ), R 1 =... = R l = m p servers. def v = (v 1,..., v k ) any fractional vertex cover; v 0 = i v i. Organize the p servers in a hypercube: [p] [p v 1 v k v0 ] [p v0 ]. Choose k independent hash functions h 1,..., h k Round 1 Each server sends each tuple R j (x j1, x j2,...) to all servers whose coordinates j 1, j 2,... are h j1 (x j1 ), h j2 (x j2 ),... and broadcasts along the missing dimensions. Then, each server computes Q on its local data. Theorem The load of the HyperCube algorithm is L = l m p 1/v 0 Proof. Fix a server. E[# tuples in R j ] = = O( m p 1/v 0 ). m p i R j v i /v 0 m p 1/v 0 since i R j v i 1. Dan Suciu Multi-Joins Lecture 2 March, / 34

70 Lower Bound for General Queries Definition R [n] r is called a matching of arity r if R = n and every column is a key. Example: matching of arity 3, n = 4 R x y z Theorem Suppose all arities are 2. For any packing u, denote L lower = m p 1/ j u j. Then, for any algorithm A with load L < L lower, it returns E[ A ] (L/L lower ) u E[ Q ] answers, where the expectation is over random matchings Proof in the section today. (Q: what about arities 1?) Dan Suciu Multi-Joins Lecture 2 March, / 34

71 Lower Bound for General Queries Corollary HyperCube algorithm is optimal. (Because min v i v i = max u j u j = τ.) Dan Suciu Multi-Joins Lecture 2 March, / 34

72 Summary of Lecture 2 Parallel query engines today compute one join at a time. Issues: communication rounds, intermediate results, skew. The HyperCube algorithm: one round. Restrictions: Equal-cardinalities (will generalize in Lecture 3) Skew-free databases (skew is open, but will discuss in Lecture 4). HyperCube is more resilient to skew than a join. A surprising fact: Parallel algorithms: lower bound given by fractional edge packing. Sequential algorithms: lower given by fractional edge cover. Multiple rounds: all lower bounds [Beame 13] use weaker model. Open problem: an algorithm with load O(n/p) cannot compute Q = R 1 (x 0, x 1 ), R 2 (x 1, x 2 ), R 3 (x 2, x 3 ), R 4 (x 3, x 4 ), R 5 (x 4, x 5 ) in two rounds, over random permutations. Dan Suciu Multi-Joins Lecture 2 March, / 34