Advanced Query Processing
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1 Advanced Query Processing CSEP544 Nov CSEP544 Optimal Sequential Algorithms Nov / 28
2 Lecture 9: Advanced Query Processing Optimal Sequential Algorithms. Semijoin Reduction Optimal Parallel Algorithms. CSEP544 Optimal Sequential Algorithms Nov / 28
3 Bibliography E Friedgut, Hypergraphs, entropy, and inequalities, American Mathematical Monthly, , Albert Atserias, Martin Grohe, Dniel Marx: Size Bounds and Query Plans for Relational Joins. SIAM J. Comput. 42(4): (2013) Hung Q. Ngo, Christopher Ré, Atri Rudra: Skew strikes back: new developments in the theory of join algorithms. SIGMOD Record 42(4): 5-16 (2013) Paul Beame, Paraschos Koutris, Dan Suciu: Skew in parallel query processing. PODS 2014: Paul Beame, Paraschos Koutris, Dan Suciu: Communication steps for parallel query processing. PODS 2013: CSEP544 Optimal Sequential Algorithms Nov / 28
4 Natural Join R S Joins R, S on all common attributes, removes duplicate attributes R A B a 1 b 1 a 1 b 2 a 2 b 3 a 3 b 4 S A C a 1 c 1 a 1 c 2 a 3 c 3 a 4 c 4 T = R S A B C a 1 b 1 c 1 a 1 b 1 c 2 a 1 b 2 c 1 a 1 b 2 c 2 a 3 b 4 c 3 Input schemas: R(A, B), S(A, C) Output schema: T (A, B, C) CSEP544 Optimal Sequential Algorithms Nov / 28
5 Very Quick Review of Basic Join Algorithms Compute R A=B S Nested-loop join Hash-join Merge-join (To describe in class.) Complexity: O(( R + S + R A=B S ) log( R + S )) Ignoring log factors, Complexity: O( Input + Output ) CSEP544 Optimal Sequential Algorithms Nov / 28
6 Conjunctive Queries Example Q 1 (x, y, z, u) = R(x, y), S(y, z), T (z, u) Relational Algebra: (R(x, y) S(y, z)) T (z, u) First Order Logic: Q 1 = {(x, y, z, u) (x, y) R (y, z) S (z, u) T } SQL: select from where CSEP544 Optimal Sequential Algorithms Nov / 28
7 Conjunctive Queries Example Q 1 (x, y, z, u) = R(x, y), S(y, z), T (z, u) Relational Algebra: (R(x, y) S(y, z)) T (z, u) First Order Logic: Q 1 = {(x, y, z, u) (x, y) R (y, z) S (z, u) T } SQL: select from where Example Q 2 (x, u) = R(x, y), S(y, z), T (z, u) Relational Algebra: Π x,u ((R(x, y) S(y, z)) T (z, u)) First Order Logic: Q 1 = {(x, u) y z((x, y) R (y, z) S (z, u) T )} SQL: select from where CSEP544 Optimal Sequential Algorithms Nov / 28
8 Traditional Approach to Computing Conjunctive Queries Q(x, y, z) = R(x, y), S(y, z), T (z, x) Optimizer generates a query plan: Temp(x, y, z) =R(x, y) S(y, z) Q(x, y, z) =Temp(x, y, z) T (z, x) Optimizers examines many possible plans, evaluates the cheapest plan. Problem: intermediate results may be large, and very hard to estimate. CSEP544 Optimal Sequential Algorithms Nov / 28
9 Upper Bound on the Size of the Answer Consider the join of two relations: Q(x, y, z) = R(x, y), S(y, z) Question If R = m 1, S = m 2, how large can Q be? CSEP544 Optimal Sequential Algorithms Nov / 28
10 Upper Bound on the Size of the Answer Consider the join of two relations: Q(x, y, z) = R(x, y), S(y, z) Question If R = m 1, S = m 2, how large can Q be? Can be 0 CSEP544 Optimal Sequential Algorithms Nov / 28
11 Upper Bound on the Size of the Answer Consider the join of two relations: Q(x, y, z) = R(x, y), S(y, z) Question If R = m 1, S = m 2, how large can Q be? Can be 0 Can be m 1 m 2 CSEP544 Optimal Sequential Algorithms Nov / 28
12 Upper Bound on the Size of the Answer Consider the join of two relations: Q(x, y, z) = R(x, y), S(y, z) Question If R = m 1, S = m 2, how large can Q be? Can be 0 Can be m 1 m 2 Answer: 0 Q m 1 m 2. CSEP544 Optimal Sequential Algorithms Nov / 28
13 Upper Bound on the Size of the Answer Q(x, y, z) = R(x, y), S(y, z), T (z, x) Question If R = m 1, S = m 2, T = m 3, how large can the result be? CSEP544 Optimal Sequential Algorithms Nov / 28
14 Upper Bound on the Size of the Answer Q(x, y, z) = R(x, y), S(y, z), T (z, x) Question If R = m 1, S = m 2, T = m 3, how large can the result be? Naive answer: m 1 m 2 m 3 (why?) CSEP544 Optimal Sequential Algorithms Nov / 28
15 Upper Bound on the Size of the Answer Q(x, y, z) = R(x, y), S(y, z), T (z, x) Question If R = m 1, S = m 2, T = m 3, how large can the result be? Naive answer: m 1 m 2 m 3 (why?) Better answer: m 1 m 2 (why?) CSEP544 Optimal Sequential Algorithms Nov / 28
16 Upper Bound on the Size of the Answer Q(x, y, z) = R(x, y), S(y, z), T (z, x) Question If R = m 1, S = m 2, T = m 3, how large can the result be? Naive answer: m 1 m 2 m 3 (why?) Better answer: m 1 m 2 (why?) But also: m 1 m 3, m 2 m 3 CSEP544 Optimal Sequential Algorithms Nov / 28
17 Upper Bound on the Size of the Answer Q(x, y, z) = R(x, y), S(y, z), T (z, x) Question If R = m 1, S = m 2, T = m 3, how large can the result be? Naive answer: m 1 m 2 m 3 (why?) Better answer: m 1 m 2 (why?) But also: m 1 m 3, m 2 m 3 We will show: Also (and better!): Q m 1 m 2 m 3 There exists an algorithm that computes Q in time min(all the above) How this generalizes to any conjunctive query. CSEP544 Optimal Sequential Algorithms Nov / 28
18 The Hypergraph of a Query Definition Let Q be a full conjunctive query without self-joins. The hypergraph G of Q consists of: Nodes(G) = Vars(Q) the set of variables of Q HyperEdges(G) = Atoms(Q) the set of atoms of Q. CSEP544 Optimal Sequential Algorithms Nov / 28
19 The Hypergraph of a Query Definition Let Q be a full conjunctive query without self-joins. The hypergraph G of Q consists of: Nodes(G) = Vars(Q) the set of variables of Q HyperEdges(G) = Atoms(Q) the set of atoms of Q. Q(x,y,z) = R(x,y),S(y,z),T(z,x) x z y Q(x,y,z) = R(x,y,z),S(x),T(y),K(z),M(x,u) u x z y CSEP544 Optimal Sequential Algorithms Nov / 28
20 Edge Cover of a Hypergraph G G = nodes x 1,..., x k and hyperedges R 1,..., R l. Definition An edge cover = subset of edges that contain all nodes. Full conjunctive query: Q(x) = R 1 (x 1 ),..., R l (x l ) Relation sizes: R 1 = m 1,..., R l = m l Proposition (Simple!) Let R i1,..., R iu be any edge cover. Then Q m i1 m i2 m iu (proof in class) CSEP544 Optimal Sequential Algorithms Nov / 28
21 Edge Cover of a Hypergraph G G = nodes x 1,..., x k and hyperedges R 1,..., R l. Definition An edge cover = subset of edges that contain all nodes. Full conjunctive query: Q(x) = R 1 (x 1 ),..., R l (x l ) Relation sizes: R 1 = m 1,..., R l = m l Proposition (Simple!) Let R i1,..., R iu be any edge cover. Then Q m i1 m i2 m iu (proof in class) CSEP544 Optimal Sequential Algorithms Nov / 28
22 Edge Cover of a Hypergraph G G = nodes x 1,..., x k and hyperedges R 1,..., R l. Definition An edge cover = subset of edges that contain all nodes. Full conjunctive query: Q(x) = R 1 (x 1 ),..., R l (x l ) Relation sizes: R 1 = m 1,..., R l = m l Proposition (Simple!) Let R i1,..., R iu be any edge cover. Then Q m i1 m i2 m iu (proof in class) CSEP544 Optimal Sequential Algorithms Nov / 28
23 Edge Cover of a Hypergraph G G = nodes x 1,..., x k and hyperedges R 1,..., R l. Definition An edge cover = subset of edges that contain all nodes. Full conjunctive query: Q(x) = R 1 (x 1 ),..., R l (x l ) Relation sizes: R 1 = m 1,..., R l = m l Proposition (Simple!) Let R i1,..., R iu be any edge cover. Then Q m i1 m i2 m iu (proof in class) CSEP544 Optimal Sequential Algorithms Nov / 28
24 Fractional Edge Cover of a Hypergraph G G = nodes x 1,..., x k and hyperedges R 1,..., R l. Definition A fractional edge cover = sequence of positive numbers u 1,..., u l s.t.: i j x i R j u j 1 Theorem (AGM 13) Let u 1,..., u l be any fractional edge cover. Then Q m u 1 1 mu 2 2 mu l l CSEP544 Optimal Sequential Algorithms Nov / 28
25 Fractional Edge Cover of a Hypergraph G G = nodes x 1,..., x k and hyperedges R 1,..., R l. Definition A fractional edge cover = sequence of positive numbers u 1,..., u l s.t.: i j x i R j u j 1 Theorem (AGM 13) Let u 1,..., u l be any fractional edge cover. Then Q m u 1 1 mu 2 2 mu l l CSEP544 Optimal Sequential Algorithms Nov / 28
26 Fractional Edge Cover of a Hypergraph G G = nodes x 1,..., x k and hyperedges R 1,..., R l. Definition A fractional edge cover = sequence of positive numbers u 1,..., u l s.t.: i j x i R j u j 1 Theorem (AGM 13) Let u 1,..., u l be any fractional edge cover. Then Q m u 1 1 mu 2 2 mu l l CSEP544 Optimal Sequential Algorithms Nov / 28
27 Examples Q(x, y, z) = R(x, y), S(y, z), T (z, x) R = S = T = m AGM u (Q) = m u 1 1 mu 2 2 mu l l CSEP544 Optimal Sequential Algorithms Nov / 28
28 Examples Q(x, y, z) = R(x, y), S(y, z), T (z, x) R = S = T = m AGM u (Q) = m u 1 1 mu 2 2 mu l l A fractional edge: u = (1/2, 1/2, 1/2) 1/2 x 1/2 z 1/2 y CSEP544 Optimal Sequential Algorithms Nov / 28
29 Examples Q(x, y, z) = R(x, y), S(y, z), T (z, x) R = S = T = m AGM u (Q) = m u 1 1 mu 2 2 mu l l A fractional edge: u = (1/2, 1/2, 1/2) 1/2 x 1/2 z 1/2 y It follows that Q m 1/2 m 1/2 m 1/2 = m 3/2 With m edges you can built at most m 3/2 triangles! CSEP544 Optimal Sequential Algorithms Nov / 28
30 AGM Bound Definition AGM(Q) = min u m u 1 1 mu 2 2 mu l l Thus: Q AGM(Q). Example Q(x, y, z) = R(x, y), S(y, z), T (z, x), R = m 1, S = m 2, T = m 3 u = (1, 1, 0) (1, 0, 1) (0, 1, 1) ( 1 2, 1 2, 1 2 ) AGM(Q) = min of m 1 m 2 m 1 m 3 m 2 m 3 (m 1 m 2 m 3 ) 1/2 Example Q(x, y, z, v, w) = R(x, y), S(y, z), T (z, v), K(v, w) u = (1, 0, 1, 1) (1, 1, 0, 1) AGM(Q) = min of m 1 m 3 m 4 m 1 m 2 m 4 CSEP544 Optimal Sequential Algorithms Nov / 28
31 AGM Bound Definition AGM(Q) = min u m u 1 1 mu 2 2 mu l l Thus: Q AGM(Q). Example Q(x, y, z) = R(x, y), S(y, z), T (z, x), R = m 1, S = m 2, T = m 3 u = (1, 1, 0) (1, 0, 1) (0, 1, 1) ( 1 2, 1 2, 1 2 ) AGM(Q) = min of m 1 m 2 m 1 m 3 m 2 m 3 (m 1 m 2 m 3 ) 1/2 Example Q(x, y, z, v, w) = R(x, y), S(y, z), T (z, v), K(v, w) u = (1, 0, 1, 1) (1, 1, 0, 1) AGM(Q) = min of m 1 m 3 m 4 m 1 m 2 m 4 CSEP544 Optimal Sequential Algorithms Nov / 28
32 AGM Bound Definition AGM(Q) = min u m u 1 1 mu 2 2 mu l l Thus: Q AGM(Q). Example Q(x, y, z) = R(x, y), S(y, z), T (z, x), R = m 1, S = m 2, T = m 3 u = (1, 1, 0) (1, 0, 1) (0, 1, 1) ( 1 2, 1 2, 1 2 ) AGM(Q) = min of m 1 m 2 m 1 m 3 m 2 m 3 (m 1 m 2 m 3 ) 1/2 Example Q(x, y, z, v, w) = R(x, y), S(y, z), T (z, v), K(v, w) u = (1, 0, 1, 1) (1, 1, 0, 1) AGM(Q) = min of m 1 m 3 m 4 m 1 m 2 m 4 CSEP544 Optimal Sequential Algorithms Nov / 28
33 The Worst-Case Query Output Question Can the query output ever get as large as the AGM bound? AGM u (Q) = m u 1 1 mu 2 2 mu l l CSEP544 Optimal Sequential Algorithms Nov / 28
34 The Worst-Case Query Output Question Can the query output ever get as large as the AGM bound? AGM u (Q) = m u 1 1 mu 2 2 mu l l Example Q(x, y, z) = R(x, y), S(y, z), T (z, x) Find three relations R = S = T = m such that Q = m 3/2 CSEP544 Optimal Sequential Algorithms Nov / 28
35 The Worst-Case Query Output Question Can the query output ever get as large as the AGM bound? AGM u (Q) = m u 1 1 mu 2 2 mu l l Example Q(x, y, z) = R(x, y), S(y, z), T (z, x) Find three relations R = S = T = m such that Q = m 3/2 Answer: let n = m 1/2, and R = S = T = [n] [n]. Then Q = n 3 = m 3/2. CSEP544 Optimal Sequential Algorithms Nov / 28
36 Background in Algebra: Duality of Linear Programs AGM u (Q) = m u 1 1 mu 2 2 mu l l AGM(Q) = min u AGM u (Q) is the optimal solution to: minimize u j log m j j i j x i R j u j 1 CSEP544 Optimal Sequential Algorithms Nov / 28
37 Background in Algebra: Duality of Linear Programs AGM u (Q) = m u 1 1 mu 2 2 mu l l AGM(Q) = min u AGM u (Q) is the optimal solution to: minimize u j log m j j i j x i R j u j 1 Fractional Edge Cover maximize v i i j v i log m j i x i R j Fractional Vertex Packing CSEP544 Optimal Sequential Algorithms Nov / 28
38 Background in Algebra: Duality of Linear Programs AGM u (Q) = m u 1 1 mu 2 2 mu l l AGM(Q) = min u AGM u (Q) is the optimal solution to: minimize u j log m j j i j x i R j u j 1 Fractional Edge Cover maximize v i i j v i log m j i x i R j Fractional Vertex Packing Theorem (Strong Duality of Linear Programs) min u j u j log m j = max v v i i CSEP544 Optimal Sequential Algorithms Nov / 28
39 Background in Algebra: Duality of Linear Programs Q(x, y, z) = R(x, y), S(y, z), T (z, x) 1/2 x 1/2 z 1/2 y min(u R log R + u S log S + u T log T ) x u R + u T 1 y u R + u S 1 z u S + u T 1 CSEP544 Optimal Sequential Algorithms Nov / 28
40 Background in Algebra: Duality of Linear Programs Q(x, y, z) = R(x, y), S(y, z), T (z, x) 1/2 x 1/2 z 1/2 y min(u R log R + u S log S + u T log T ) x u R + u T 1 y u R + u S 1 z u S + u T 1 max(v x + v y + v z) R v x + v y log R S v y + v z log S T v x + v z log T Strong duality theorem: min u R log R + u S log S + u T log T = max v x + v y + v z. CSEP544 Optimal Sequential Algorithms Nov / 28
41 Background in Algebra: Duality of Linear Programs Q(x, y, z) = R(x, y), S(y, z), T (z, x) 1/2 x 1/2 z 1/2 y min(u R log R + u S log S + u T log T ) x u R + u T 1 y u R + u S 1 z u S + u T 1 max(v x + v y + v z) R v x + v y log R S v y + v z log S T v x + v z log T Strong duality theorem: min u R log R + u S log S + u T log T = max v x + v y + v z. In class: what are the optimal solutions in these cases: R = S = T R = S T CSEP544 Optimal Sequential Algorithms Nov / 28
42 The AGM Bound is Tight AGM u (Q) = m u 1 1 mu 2 2 mu l l AGM(Q) = min u AGM u (Q) is the optimal solution to: minimize u j log m j j i j x i R j u j 1 maximize v i i j v i log m j i x i R j Theorem The AGM bound is tight Proof: start with an optimal solution v i. Define R(x 1, x 5, x 8 ) = [2 v 1 ] [2 v 5 ] [2 v 8 ] etc Then Q = 2 v 1+v = 2 u 1 log m 1 +u 2 log m = m u 1 1 mu 2 2 CSEP544 Optimal Sequential Algorithms Nov / 28
43 Computing Full Conjunctive Queries Recall: all database systems compute one join at a time This may be much larger than the maximum output size, AGM(Q). Goal: design a algorithm that runs in time AGM(Q). Worst-Case-Optimal algorithm: runs in time AGM(Q). CSEP544 Optimal Sequential Algorithms Nov / 28
44 Generic Join Compute Q(x) = R 1 (x 1 ),..., R l (x l ) If x = 1 then return R 1 R l. Otherwise, choose a variable x Assuem it occurrs in atoms R i1,..., R ik Compute A = Π x (R i1 ) Π x (R ik ) For each a A, compute Result a = Q[a/x] using Generic-Join Return a Result a. Runtime: O(AGM(Q)) (Plus a log n factor for index lookup) CSEP544 Optimal Sequential Algorithms Nov / 28
45 Generic Join Compute Q(x) = R 1 (x 1 ),..., R l (x l ) If x = 1 then return R 1 R l. Otherwise, choose a variable x Assuem it occurrs in atoms R i1,..., R ik Compute A = Π x (R i1 ) Π x (R ik ) For each a A, compute Result a = Q[a/x] using Generic-Join Return a Result a. Runtime: O(AGM(Q)) (Plus a log n factor for index lookup) CSEP544 Optimal Sequential Algorithms Nov / 28
46 Generic Join Example Q(x, y, z) = R(x, y), S(y, z), T (z, x) Compute A = Π x (R) Π x (T ) = {a 1,..., a n } For each a i A, denote R (y) = R(a i, y), T (z) = T (z, a i ) Compute Result i (a i, y, z) = R (y), S(y, z), T (z) Return i Result i Runtime: O(m 3/2 ) assuming R = S = T = m. CSEP544 Optimal Sequential Algorithms Nov / 28
47 Details of Generic Join Fix variable order: x 1, x 2,... (AGM bound always holds, but in practice the order can matter a lot) Order/index the relations accordingly. For example, R(x 3, x 6, x 7 ) has a B+-index on x 3, x 6, x 7. Computing A = Π x (R i1 ) Π x (R ik ) is similar to multi-way merge join. Must ensure that runtime is min( R i1, R i2,...). When we iterate a A, we are making one more binding in all indexes. LeapFrog Tree Join (by LogicBlox) is based on these ideas. CSEP544 Optimal Sequential Algorithms Nov / 28
48 Details of Generic Join Q(x, y, z) = R(x, y), S(y, z), T (z, x) Compute A = Π x (R) Π x (T ) = {a 1,..., a n } For each a i A, denote R (y) = R(a i, y), T (z) = T (z, a i ) Compute Result i (a i, y, z) = R (y), S(y, z), T (z) Return i Result i Runtime: O(m 3/2 ) assuming R = S = T = m. Variable order: x, y, z. What happens in each case? Complete bipartite graph: R = S = T = [m 1/2 ] [m 1/2 ]. Skewed at x: R = [1] [m], S [m] [m], T = [m] [1]. Skewed at y: R = [m] [1], S = [1] [m], T [m] [m] Skewed at z: R [m] [m], S = [m] [1], T [1] [m] CSEP544 Optimal Sequential Algorithms Nov / 28
49 Discussion Major advantage over one-join-at-at-time algorithms for cyclic queries! For acyclic queries, the story is more complex (discussed later) We haven t proven its optimality yet: will do this next. CSEP544 Optimal Sequential Algorithms Nov / 28
50 Friedgut s Inequality Cauchy-Schwartz: i a i b i ( i a 2 i ) 1 2 ( i b 2 i ) 1 2 CSEP544 Optimal Sequential Algorithms Nov / 28
51 Friedgut s Inequality Cauchy-Schwartz: i a i b i ( i a 2 i ) 1 2 ( i b 2 i ) 1 2 Triangle: i,j,k a ij b jk c ki ( i,j a 2 ij ) 1 2 ( j,k b 2 jk ) 1 2 ( k,i c 2 ki ) 1 2 CSEP544 Optimal Sequential Algorithms Nov / 28
52 Friedgut s Inequality Cauchy-Schwartz: i a i b i ( i a 2 i ) 1 2 ( i b 2 i ) 1 2 Triangle: i,j,k a ij b jk c ki ( i,j a 2 ij ) 1 2 ( j,k b 2 jk ) 1 2 ( k,i c 2 ki ) 1 2 Hölder (u + v + w 1): i a i b i c i ( i a 1 u i ) u ( i b 1 v i ) v ( i c 1 w i ) w CSEP544 Optimal Sequential Algorithms Nov / 28
53 Friedgut s Inequality Cauchy-Schwartz: i a i b i ( i a 2 i ) 1 2 ( i b 2 i ) 1 2 Triangle: i,j,k a ij b jk c ki ( i,j a 2 ij ) 1 2 ( j,k b 2 jk ) 1 2 ( k,i c 2 ki ) 1 2 Hölder (u + v + w 1): Theorem (Friedgut 04) i a i b i c i ( i a 1 u i ) u ( i b 1 v i ) v ( i c 1 w i Let Q(x) = R 1 (x 1 ),..., R l (x l ) be a query and u 1,..., u l be a fractional edge cover. Then: 1 u 1 1 u l u x a 1,x1 a l,xl ( x1 a 1 ( a u l xl 1,x 1 ) l,x l ) ) w What are the queries in the examples above? CSEP544 Optimal Sequential Algorithms Nov / 28
54 Friedgut s Inequality Cauchy-Schwartz: i a i b i ( i a 2 i ) 1 2 ( i b 2 i ) 1 2 Triangle: i,j,k a ij b jk c ki ( i,j a 2 ij ) 1 2 ( j,k b 2 jk ) 1 2 ( k,i c 2 ki ) 1 2 Hölder (u + v + w 1): Theorem (Friedgut 04) i a i b i c i ( i a 1 u i ) u ( i b 1 v i ) v ( i c 1 w i Let Q(x) = R 1 (x 1 ),..., R l (x l ) be a query and u 1,..., u l be a fractional edge cover. Then: 1 u 1 1 u l u x a 1,x1 a l,xl ( x1 a 1 ( a u l xl 1,x 1 ) l,x l ) ) w What are the queries in the examples above? Q Cauchy-Schwartz (x) = R(x), S(x); CSEP544 Optimal Sequential Algorithms Nov / 28
55 Friedgut s Inequality Cauchy-Schwartz: i a i b i ( i a 2 i ) 1 2 ( i b 2 i ) 1 2 Triangle: i,j,k a ij b jk c ki ( i,j a 2 ij ) 1 2 ( j,k b 2 jk ) 1 2 ( k,i c 2 ki ) 1 2 Hölder (u + v + w 1): Theorem (Friedgut 04) i a i b i c i ( i a 1 u i ) u ( i b 1 v i ) v ( i c 1 w i Let Q(x) = R 1 (x 1 ),..., R l (x l ) be a query and u 1,..., u l be a fractional edge cover. Then: 1 u 1 1 u l u x a 1,x1 a l,xl ( x1 a 1 ( a u l xl 1,x 1 ) l,x l ) ) w What are the queries in the examples above? Q Cauchy-Schwartz (x) = R(x), S(x); Q triangle (x, y, z) = R(x, y), S(y, z), T (z, x); CSEP544 Optimal Sequential Algorithms Nov / 28
56 What are the queries in the examples above? Q Cauchy-Schwartz (x) = R(x), S(x); Q triangle (x, y, z) = R(x, y), S(y, z), T (z, x); Q Hölder (x) = R(x), S(x), T (x) CSEP544 Optimal Sequential Algorithms Nov / 28 Friedgut s Inequality Cauchy-Schwartz: i a i b i ( i a 2 i ) 1 2 ( i b 2 i ) 1 2 Triangle: i,j,k a ij b jk c ki ( i,j a 2 ij ) 1 2 ( j,k b 2 jk ) 1 2 ( k,i c 2 ki ) 1 2 Hölder (u + v + w 1): Theorem (Friedgut 04) i a i b i c i ( i a 1 u i ) u ( i b 1 v i ) v ( i c 1 w i Let Q(x) = R 1 (x 1 ),..., R l (x l ) be a query and u 1,..., u l be a fractional edge cover. Then: 1 u 1 1 u l u x a 1,x1 a l,xl ( x1 a 1 ( a u l xl 1,x 1 ) l,x l ) ) w
57 Friedgut s Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l x a u 1 1,x 1 a u l l,x l ( x1 a 1,x1 ) u1 ( xl a l,xl ) u l CSEP544 Optimal Sequential Algorithms Nov / 28
58 Friedgut s Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l x a u 1 1,x 1 a u l l,x l ( x1 a 1,x1 ) u1 ( xl a l,xl ) u l Proof: by induction on x CSEP544 Optimal Sequential Algorithms Nov / 28
59 Friedgut s Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l x a u 1 1,x 1 a u l l,x l ( x1 a 1,x1 ) u1 ( xl a l,xl ) u l Proof: by induction on x Base Case. x = 1: Q(x) = R 1 (x),..., R l (x), u u l 1 Prove: x a u1 1,x au l l,x ( x a 1,x ) u1 ( x a l,x ) u l This is Hölder. CSEP544 Optimal Sequential Algorithms Nov / 28
60 Friedgut s Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l x a u 1 1,x 1 a u l l,x l ( x1 a 1,x1 ) u1 ( xl a l,xl ) u l Proof: by induction on x Base Case. x = 1: Q(x) = R 1 (x),..., R l (x), u u l 1 Prove: x a u1 1,x au l l,x ( x a 1,x ) u1 ( x a l,x ) u l This is Hölder. Induction Step. Pick a variable x, and remove it. For example, Q(x, y, z) = R(x, y), S(y, z), T (z, x) becomes Q (y, z) = R (y), S(y, z), T (z) CSEP544 Optimal Sequential Algorithms Nov / 28
61 Friedgut s Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l x a u 1 1,x 1 a u l l,x l ( x1 a 1,x1 ) u1 ( xl a l,xl ) u l Proof: by induction on x Base Case. x = 1: Q(x) = R 1 (x),..., R l (x), u u l 1 Prove: x a u1 1,x au l l,x ( x a 1,x ) u1 ( x a l,x ) u l This is Hölder. Induction Step. Pick a variable x, and remove it. For example, Q(x, y, z) = R(x, y), S(y, z), T (z, x) becomes Q (y, z) = R (y), S(y, z), T (z) axy u1 byzc u2 zx u3 = xyz byz u2 yz x axy u1 czx u3 group by x CSEP544 Optimal Sequential Algorithms Nov / 28
62 Friedgut s Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l x a u 1 1,x 1 a u l l,x l ( x1 a 1,x1 ) u1 ( xl a l,xl ) u l Proof: by induction on x Base Case. x = 1: Q(x) = R 1 (x),..., R l (x), u u l 1 Prove: x a u1 1,x au l l,x ( x a 1,x ) u1 ( x a l,x ) u l This is Hölder. Induction Step. Pick a variable x, and remove it. For example, Q(x, y, z) = R(x, y), S(y, z), T (z, x) becomes Q (y, z) = R (y), S(y, z), T (z) axy u1 byzc u2 zx u3 = xyz b u2 yz yz( x byz u2 yz x axy u1 czx u3 group by x a xy ) u1 ( c zx ) u3 Hölder u 1 + u 3 1 x CSEP544 Optimal Sequential Algorithms Nov / 28
63 Friedgut s Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l x a u 1 1,x 1 a u l l,x l ( x1 a 1,x1 ) u1 ( xl a l,xl ) u l Proof: by induction on x Base Case. x = 1: Q(x) = R 1 (x),..., R l (x), u u l 1 Prove: x a u1 1,x au l l,x ( x a 1,x ) u1 ( x a l,x ) u l This is Hölder. Induction Step. Pick a variable x, and remove it. For example, Q(x, y, z) = R(x, y), S(y, z), T (z, x) becomes Q (y, z) = R (y), S(y, z), T (z) axy u1 byzc u2 zx u3 = xyz b u2 yz yz( x = byza u2 u1 y C u3 yz byz u2 yz x axy u1 czx u3 group by x a xy ) u1 ( c zx ) u3 Hölder u 1 + u 3 1 x z ( yz b yz ) u2 ( y A y ) u1 ( C z ) u3 Induction for Q z CSEP544 Optimal Sequential Algorithms Nov / 28
64 Friedgut s Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l x a u 1 1,x 1 a u l l,x l ( x1 a 1,x1 ) u1 ( xl a l,xl ) u l Proof: by induction on x Base Case. x = 1: Q(x) = R 1 (x),..., R l (x), u u l 1 Prove: x a u1 1,x au l l,x ( x a 1,x ) u1 ( x a l,x ) u l This is Hölder. Induction Step. Pick a variable x, and remove it. For example, Q(x, y, z) = R(x, y), S(y, z), T (z, x) becomes Q (y, z) = R (y), S(y, z), T (z) axy u1 byzc u2 zx u3 = xyz b u2 yz yz( x = byza u2 u1 y C u3 yz =( yz b yz ) u2 ( xy byz u2 yz x axy u1 czx u3 group by x a xy ) u1 ( c zx ) u3 Hölder u 1 + u 3 1 x z ( yz b yz ) u2 ( y a xy ) u1 ( c zx ) u3 zx A y ) u1 ( C z ) u3 Induction for Q z CSEP544 Optimal Sequential Algorithms Nov / 28
65 Friedgut s Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l x a u 1 1,x 1 a u l l,x l ( x1 a 1,x1 ) u1 ( xl a l,xl ) u l Proof: by induction on x Base Case. x = 1: Q(x) = R 1 (x),..., R l (x), u u l 1 Prove: x a u1 1,x au l l,x ( x a 1,x ) u1 ( x a l,x ) u l This is Hölder. Induction Step. Pick a variable x, and remove it. For example, Q(x, y, z) = R(x, y), S(y, z), T (z, x) becomes Q (y, z) = R (y), S(y, z), T (z) axy u1 byzc u2 zx u3 = xyz b u2 yz yz( x = byza u2 u1 y C u3 yz =( yz b yz ) u2 ( xy byz u2 yz x axy u1 czx u3 group by x a xy ) u1 ( c zx ) u3 Hölder u 1 + u 3 1 x z ( yz b yz ) u2 ( y a xy ) u1 ( c zx ) u3 zx A y ) u1 ( C z ) u3 Induction for Q z QED CSEP544 Optimal Sequential Algorithms Nov / 28
66 The AGM Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l Sizes R 1 = m 1,..., R l = m l Prove Q m u 1 1 mu l l Let Dom = the domain of all constants in the relations R 1,..., R l. For every j = 1,..., l, and every tuple x j Dom x j, define: 1 if the tuple x j belongs to R j a j,xj = 0 otherwise Then: m j = R j = xj Dom x j a j,x j, Q = x Dom x a 1,x1 a l,xl Now use Friedgut s inequality: Q = a u 1 1,x 1 a u l l,x l ( a 1,x1 ) u 1 ( a l,xl ) u l = m u 1 1 mu l l x x 1 x l QED CSEP544 Optimal Sequential Algorithms Nov / 28
67 The AGM Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l Sizes R 1 = m 1,..., R l = m l Prove Q m u 1 1 mu l l Let Dom = the domain of all constants in the relations R 1,..., R l. For every j = 1,..., l, and every tuple x j Dom x j, define: 1 if the tuple x j belongs to R j a j,xj = 0 otherwise Then: m j = R j = xj Dom x j a j,x j, Q = x Dom x a 1,x1 a l,xl Now use Friedgut s inequality: Q = a u 1 1,x 1 a u l l,x l ( a 1,x1 ) u 1 ( a l,xl ) u l = m u 1 1 mu l l x x 1 x l QED CSEP544 Optimal Sequential Algorithms Nov / 28
68 The AGM Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l Sizes R 1 = m 1,..., R l = m l Prove Q m u 1 1 mu l l Let Dom = the domain of all constants in the relations R 1,..., R l. For every j = 1,..., l, and every tuple x j Dom x j, define: 1 if the tuple x j belongs to R j a j,xj = 0 otherwise Then: m j = R j = xj Dom x j a j,x j, Q = x Dom x a 1,x1 a l,xl Now use Friedgut s inequality: Q = a u 1 1,x 1 a u l l,x l ( a 1,x1 ) u 1 ( a l,xl ) u l = m u 1 1 mu l l x x 1 x l QED CSEP544 Optimal Sequential Algorithms Nov / 28
69 The AGM Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l Sizes R 1 = m 1,..., R l = m l Prove Q m u 1 1 mu l l Let Dom = the domain of all constants in the relations R 1,..., R l. For every j = 1,..., l, and every tuple x j Dom x j, define: 1 if the tuple x j belongs to R j a j,xj = 0 otherwise Then: m j = R j = xj Dom x j a j,x j, Q = x Dom x a 1,x1 a l,xl Now use Friedgut s inequality: Q = a u 1 1,x 1 a u l l,x l ( a 1,x1 ) u 1 ( a l,xl ) u l = m u 1 1 mu l l x x 1 x l QED CSEP544 Optimal Sequential Algorithms Nov / 28
70 The AGM Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l Sizes R 1 = m 1,..., R l = m l Prove Q m u 1 1 mu l l Let Dom = the domain of all constants in the relations R 1,..., R l. For every j = 1,..., l, and every tuple x j Dom x j, define: 1 if the tuple x j belongs to R j a j,xj = 0 otherwise Then: m j = R j = xj Dom x j a j,x j, Q = x Dom x a 1,x1 a l,xl Now use Friedgut s inequality: Q = a u 1 1,x 1 a u l l,x l ( a 1,x1 ) u 1 ( a l,xl ) u l = m u 1 1 mu l l x x 1 x l QED CSEP544 Optimal Sequential Algorithms Nov / 28
71 The AGM Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l Sizes R 1 = m 1,..., R l = m l Prove Q m u 1 1 mu l l Let Dom = the domain of all constants in the relations R 1,..., R l. For every j = 1,..., l, and every tuple x j Dom x j, define: 1 if the tuple x j belongs to R j a j,xj = 0 otherwise Then: m j = R j = xj Dom x j a j,x j, Q = x Dom x a 1,x1 a l,xl Now use Friedgut s inequality: Q = a u 1 1,x 1 a u l l,x l ( a 1,x1 ) u 1 ( a l,xl ) u l = m u 1 1 mu l l x x 1 x l QED CSEP544 Optimal Sequential Algorithms Nov / 28
72 The AGM Inequality Proof Query Q(x) = R 1 (x 1 ),..., R l (x l ), fractional cover u 1,..., u l Sizes R 1 = m 1,..., R l = m l Prove Q m u 1 1 mu l l Let Dom = the domain of all constants in the relations R 1,..., R l. For every j = 1,..., l, and every tuple x j Dom x j, define: 1 if the tuple x j belongs to R j a j,xj = 0 otherwise Then: m j = R j = xj Dom x j a j,x j, Q = x Dom x a 1,x1 a l,xl Now use Friedgut s inequality: Q = a u 1 1,x 1 a u l l,x l ( a 1,x1 ) u 1 ( a l,xl ) u l = m u 1 1 mu l l x x 1 x l QED CSEP544 Optimal Sequential Algorithms Nov / 28
73 Proof of Optimality for Generic Join Use exactly the same induction step as in Friedgut s inequality. CSEP544 Optimal Sequential Algorithms Nov / 28
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