Fundamentos lógicos de bases de datos (Logical foundations of databases)

Transcription

1 20/7/2015 ECI 2015 Buenos Aires Fundamentos lógicos de bases de datos (Logical foundations of databases) Diego Figueira Gabriele Puppis CNRS LaBRI

2 About the speakers Gabriele Puppis PhD from Udine (Italy) post-docs in Oxford Works in LaBRI, Bordeaux CNRS researcher Diego Figueira PhD from ENS Cachan (France), post-docs in Warsaw, Edinburgh Works in LaBRI, Bordeaux CNRS researcher Alma Mater: UBA! 2

3 About the speakers Gabriele Puppis PhD from Udine (Italy) post-docs in Oxford Works in LaBRI, Bordeaux CNRS researcher Diego Figueira PhD from ENS Cachan (France), post-docs in Warsaw, Edinburgh Works in LaBRI, Bordeaux CNRS researcher Alma Mater: UBA! 2

4 About the speakers Gabriele Puppis PhD from Udine (Italy) post-docs in Oxford Works in LaBRI, Bordeaux CNRS researcher hermano del inigualable Diego Figueira PhD from ENS Cachan (France), post-docs in Warsaw, Edinburgh Works in LaBRI, Bordeaux CNRS researcher Alma Mater: UBA!!!! 2

5 First and foremost interrupt! ask! (in any language) 3

6 Organization Format: 45 min slots / 20 min breaks: Schedule: Relational Algebra First-Order logic EF games, 0-1 law, Locality Conjunctive Queries logic databases complexity Evaluation: A short test on Saturday Slides available at 4

7 Databases database = a collection of data, structured in some way + a way of defining, querying, updating the data inside mediate between humans, processes & data * * [Abitebou, Hull, Vianu Foundations of databases ] 5

8 Databases database = a collection of data, structured in some way + a way of defining, querying, updating the data inside mediate between humans, processes & data * Data model how the data is logically organized mathematical abstraction for representing data independent from physical organisation DBMS also implement: transactions, concurrency, access control, resiliency * [Abitebou, Hull, Vianu Foundations of databases ] 5

9 Relational databases Relational data model = data logically organised into relations ( tables ). What s a relation? a (finite) subset of the cartesian product of sets a table with rows and columns 6

10 Relational databases Relational data model = data logically organised into relations ( tables ). What s a relation? a (finite) subset of the cartesian product of sets a table with rows and columns like: { (1,a,2), (2,b,6), (2,a,1) } N {a,b} N 6

11 Relational databases Relational data model = data logically organised into relations ( tables ). What s a relation? a (finite) subset of the cartesian product of sets a table with rows and columns like: { (1,a,2), (2,b,6), (2,a,1) } N {a,b} N a tuple (a 3-tuple ) 6

12 Relational databases Relational data model = data logically organised into relations ( tables ). What s a relation? a (finite) subset of the cartesian product of sets a table with rows and columns like: { (1,a,2), (2,b,6), (2,a,1) } N {a,b} N a tuple (a 3-tuple ) () 0-tuple 6

13 Relational databases Relational data model = data logically organised into relations ( tables ). What s a relation? a (finite) subset of the cartesian product of sets a table with rows and columns like: { (1,a,2), (2,b,6), (2,a,1) } N {a,b} N like: 1 a 2 2 b 6 2 a 1 a tuple (a 3-tuple ) () 0-tuple 6

14 Relational databases Relational data model = data logically organised into relations ( tables ). What s a relation? a (finite) subset of the cartesian product of sets a table with rows and columns DB = A schema: names of tables and attributes An instance: data conforming to the schema 6

15 Relational databases Relational data model = data logically organised into relations ( tables ). What s a relation? a (finite) subset of the cartesian product of sets a table with rows and columns DB = A schema: names of tables and attributes Films (Title:string, Director:string, Actor:string) Schedule (Theatre:string, Title:string) An instance: data conforming to the schema 6

16 Relational databases Relational data model = data logically organised into relations ( tables ). What s a relation? a (finite) subset of the cartesian product of sets a table with rows and columns DB = A schema: names of tables and attributes Films (Title:string, Director:string, Actor:string) Schedule (Theatre:string, Title:string) An instance: data conforming to the schema Films Schedule Title Director Actor Theatre Title 8 1/2 Fellini Mastroianni Utopia Dr. Strangelove Shining Kubrick Nicholson Utopia 8 1/2 Dr. Strangelove Kubrick Sellers UGC Dr. Strangelove 8 femmes Ozon Ardant UGC 8 femmes 6

17 Relational databases Relational data model = data logically organised into relations ( tables ). We assume all elements come from a fixed set of constants or data values U. 7

18 Relational databases: queries What is a query q? A mapping that takes a database instance D returns a relation q(d) U r of fixed arity r 8

19 Relational databases: queries computable! What is a query q? A mapping that takes a database instance D returns a relation q(d) U r of fixed arity r 8

20 Relational databases: queries computable! What is a query q? A mapping that takes a database instance D returns a relation q(d) U r of fixed arity r generic! (order independent) 8

21 Relational databases: queries computable! What is a query q? A mapping that takes a database instance D returns a relation q(d) U r of fixed arity r generic! (order independent) Boolean query: r=0 Either yes { () } or no { } 8

22 Relational databases: queries What is a query q? A mapping that takes a database instance D returns a relation q(d) U r of fixed arity r What do we care about queries? expressive power evaluation static analysis 8

23 The fundamental questions: How to query the relational data model? How efficient/expressive is it? 9

24 The fundamental questions: How to query the relational data model? How efficient/expressive is it? expressiveness efficiency 9

25 Query languages Query Language Syntax + Semantics Expressions for querying the db, governed by syntactic rules Select X from Y y :- x (x y) Interpretation of symbols in terms of some structure Retrieves all strings in column X of table Y Returns the maximum element of the set. 10

26 Relational Algebra (RA) [Codd, 1970] Syntax: E := R,S, E E E \ E E E π M (E) σ ϴ (E) where M N ϴ N {=, } N 11

27 Relational Algebra (RA) [Codd, 1970] Syntax: E := R,S, E E E \ E E E π M (E) σ ϴ (E) where M N ϴ N {=, } N R 1 R 2 : Set union R 1 R 2 : Cartesian product R 1 \ R 2 : Set difference 11

28 Relational Algebra (RA) [Codd, 1970] Syntax: E := R,S, E E E \ E E E π M (E) σ ϴ (E) where M N ϴ N {=, } N R 1 R 2 : Set union R 1 R 2 : Cartesian product R 1 \ R 2 : Set difference σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} : Selection 11

29 Relational Algebra (RA) [Codd, 1970] Syntax: E := R,S, E E E \ E E E π M (E) σ ϴ (E) where M N ϴ N {=, } N R 1 R 2 : Set union R 1 R 2 : Cartesian product R 1 \ R 2 : Set difference σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} : Selection 11

30 Relational Algebra (RA) [Codd, 1970] Syntax: E := R,S, E E E \ E E E π M (E) σ ϴ (E) where M N ϴ N {=, } N R 1 R 2 : Set union R 1 R 2 : Cartesian product σ {1=3,1 2} ( {(1,2,1), A procedural (2,2,2)} query ) = language {(1,2,1)} R 1 \ R 2 : Set difference σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} : Selection 11

31 Relational Algebra (RA) [Codd, 1970] Syntax: E := R,S, E E E \ E E E π M (E) σ ϴ (E) where M N ϴ N {=, } N R 1 R 2 : Set union R 1 R 2 : Cartesian product σ {1=3,1 2} ( {(1,2,1), A procedural (2,2,2)} query ) = language {(1,2,1)} R 1 \ R 2 : Set difference σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} : Selection π {i1,,i n } (R) {(x i 1,,x in ) (x 1,, x m ) R} : Projection 11

32 Relational Algebra (RA) [Codd, 1970] Syntax: E := R,S, E E E \ E E E π M (E) σ ϴ (E) where M N ϴ N {=, } N R 1 R 2 : Set union R 1 R 2 : Cartesian product σ {1=3,1 2} ( {(1,2,1), A procedural (2,2,2)} query ) = language {(1,2,1)} π {1,3} ( {(1,2,1),(2,2,2)} ) = {(1,1), (2,2)} R 1 \ R 2 : Set difference σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} : Selection π {i1,,i n } (R) {(x i 1,,x in ) (x 1,, x m ) R} : Projection 11

33 Relational Algebra (RA) [Codd, 1970] R 1 R 2 : Set union R 1 R 2 : Cartesian product R 1 \ R 2 : Set difference σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} : Selection π {i1,,i n } (R) {(x i 1,,x in ) (x 1,, x m ) R} : Projection Question 1: What is the RA expression for { (v 1,v 2 ) there are w 1 w 2 so that (v 1,w 1 ) R 1 and (v 2,w 2 ) R 2 }? Question 2: π 2 (σ 1=3 (π 2 (σ 1=3 (R 1 R 2 )) R 2 ))=? R 1 R 2 a 3 a 4 b 2 b 1 c 4 b 2 b 3 a 1 a 2 b 3 12

34 Relational Algebra (RA) [Codd, 1970] R 1 R 2 : Set union R 1 R 2 : Cartesian product R 1 \ R 2 : Set difference σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} : Selection π {i1,,i n } (R) {(x i 1,,x in ) (x 1,, x m ) R} : Projection Question 1: What is the RA expression for { (v 1,v 2 ) there are w 1 w 2 so that (v 1,w 1 ) R 1 and (v 2,w 2 ) R 2 }? Answer: π {1,3} (σ 1 3 (R 1 R 2 )) Question 2: π 2 (σ 1=3 (π 2 (σ 1=3 (R 1 R 2 )) R 2 ))=? a b c c b a a b R 1 R 2 a 3 b 2 c 4 b 3 a 2 a 4 b 1 b 2 a 1 b 3 12

35 Relational Algebra (RA) [Codd, 1970] R 1 R 2 : Set union R 1 R 2 : Cartesian product R 1 \ R 2 : Set difference σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} : Selection π {i1,,i n } (R) {(x i 1,,x in ) (x 1,, x m ) R} : Projection Question 1: What is the RA expression for { (v 1,v 2 ) there are w 1 w 2 so that (v 1,w 1 ) R 1 and (v 2,w 2 ) R 2 }? Answer: π {1,3} (σ 1 3 (R 1 R 2 )) Question 2: π 2 (σ 1=3 (π 2 (σ 1=3 (R 1 R 2 )) R 2 ))=? Answer (only one element): b a b c c b a a b R 1 R 2 a 3 b 2 c 4 b 3 a 2 a 4 b 1 b 2 a 1 b 3 12

36 RA = Basic SQL no domain-specific features, aggregation, etc Select X From R 1,,R n Where Z or not in ( ) π X ( σ Z ( R 1 R n ) ) union difference 13

37 RA = Basic SQL no domain-specific features, aggregation, etc Select X From R 1,,R n Where Z or not in ( ) π X ( σ Z ( R 1 R n ) ) union difference π 2 (σ 1 3(R 1 R 2 )) a 3 R 1 R 2 b 2 c 4 b 3 a 2 a 4 b 1 b 2 a 1 b 3 13

38 RA = Basic SQL no domain-specific features, aggregation, etc Select X From R 1,,R n Where Z or not in ( ) π X ( σ Z ( R 1 R n ) ) union difference Select R 1.2 as foo π 2 (σ 1 3(R 1 R 2 )) From R 1, R 2 a 3 Where R 1.1 R 2.1 R 1 R 2 b 2 c 4 b 3 a 2 a 4 b 1 b 2 a 1 b 3 13

39 RA = Basic SQL no domain-specific features, aggregation, etc Select X From R 1,,R n Where Z or not in ( ) π X ( σ Z ( R 1 R n ) ) union difference π 2 (σ 1 3(R 1 R 2 )) From R 1, R 2 a 3 π 2 (σ 1=3 ( R 2 )) Select R 1.2 as foo Where R 1.1 R 2.1 R 1 R 2 b 2 c 4 b 3 a 2 a 4 b 1 b 2 a 1 b 3 13

40 RA = Basic SQL no domain-specific features, aggregation, etc Select X From R 1,,R n Where Z or not in ( ) π X ( σ Z ( R 1 R n ) ) union difference Select R 1.2 as foo R 1 R 2 π 2 (σ 1 3(R 1 R 2 )) From R 1, R 2 a 3 π 2 (σ 1=3 ( R 2 )) Where R 1.1 R 2.1 a 4 b 2 b 1 c 4 b 2 b 3 a 1 a 2 b 3 13

41 RA = Basic SQL no domain-specific features, aggregation, etc Select X From R 1,,R n Where Z or not in ( ) π X ( σ Z ( R 1 R n ) ) union difference Select R 1.2 as foo R 1 R 2 π 2 (σ 1 3(R 1 R 2 )) From R 1, R 2 a 3 π 2 (σ 1=3 ( R 2 )) Where R 1.1 R 2.1 Select foo From, R 2 Where foo = R 2.2 b 2 c 4 b 3 a 2 a 4 b 1 b 2 a 1 b 3 13

42 Denotational languages Procedural Algebra How to obtain the result Logics What is the property of the result Declarative 14

43 Denotational languages { Relational Algebra Procedural operations on tables Algebra How to obtain the result Logics What is the property of the result Declarative 14

44 Denotational languages { Relational Algebra Procedural operations on tables Algebra How to obtain the result Logics What is the property of the result { First Order logic properties on mathematical structures Declarative 14

45 Denotational languages { Relational Algebra Procedural operations on tables Algebra How to obtain the result Logics What is the property of the result { First Order logic properties on mathematical structures Declarative 14

46 FO = First-Order logic 15

47 Denotational languages A structure is: A = (D, R 1,, R n, f 1, f n ) D is a non-empty set, the domain R i is an m-ary relation for some m (ie, R i D m ) f i is an n-ary function for some n (ie, f i : D n D ) 16

48 Denotational languages A structure is: A = (D, R 1,, R n, f 1, f n ) D is a non-empty set, the domain R i is an m-ary relation for some m (ie, R i D m ) f i is an n-ary function for some n (ie, f i : D n D ) A graph G = (V,E) V: nodes E V 2 : edges (binary relation) (no functions) A group, like (N,+) N: natural numbers (no relations) +: N 2 N addition (binary function) 16

49 First-order logic 17

50 First-order logic FO variables x, y, z, quantifiers:, Boolean connectives:,, A language to talk about structures Variables range over the domain Atomic formulas: R(x 1,, x m ), x=y 17

51 First-order logic FO variables x, y, z, quantifiers:, Boolean connectives:,, A graph G = (V,E) V: nodes E V 2 : edges (binary relation) (no functions) A language to talk about structures Variables range over the domain Atomic formulas: R(x 1,, x m ), x=y Language to talk about graphs Variables range over nodes Atomic formulas: E(x,y), x = y 17

52 First-order logic FO variables x, y, z, quantifiers:, Boolean connectives:,, A graph G = (V,E) V: nodes E V 2 : edges (binary relation) (no functions) A language to talk about structures Variables range over the domain Atomic formulas: R(x 1,, x m ), x=y Language to talk about graphs Variables range over nodes Atomic formulas: E(x,y), x = y Formulas: Atomic formulas + connectives + quantifiers 17

53 The node x has at least two neighbours y z ( (y=z) E(x,y) E(x,z)) 18

54 The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) free x is free = not quantified (a property of a node in the graph) 18

55 The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) free x is free = not quantified (a property of a node in the graph) Each node has at least two neighbours x y z ( (y=z) E(x,y) E(x,z)) 18

56 The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) free x is free = not quantified (a property of a node in the graph) Each node has at least two neighbours ψ = x y z ( (y=z) E(x,y) E(x,z)) the formula is a sentence = no free variables (a property of the graph) 18

57 The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) free x is free = not quantified (a property of a node in the graph) Each node has at least two neighbours ψ = x y z ( (y=z) E(x,y) E(x,z)) the formula is a sentence = no free variables (a property of the graph) Question: How to express in FO Every two neighbours have a common neighbour? Does it have free variables? Is it a sentence? 18

58 The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) free x is free = not quantified (a property of a node in the graph) Each node has at least two neighbours ψ = x y z ( (y=z) E(x,y) E(x,z)) the formula is a sentence = no free variables (a property of the graph) Question: How to express in FO Every two neighbours have a common neighbour? Does it have free variables? Is it a sentence? Answer: x y ( E(x,y) z ( (E(x,z) E(z,x)) (E(y,z) E(z,y)) ) ) 18

59 Binding To evaluate a formula φ we need a graph G=(V,E) and a binding α that maps free variables of φ to nodes of G. G α φ(x 1,,x n ) α : {x 1,,x n } V assigns nodes to free variables 19

60 Binding To evaluate a formula φ we need a graph G=(V,E) and a binding α that maps free variables of φ to nodes of G. G,α satisfy φ φ is satisfiable G α φ(x 1,,x n ) α : {x 1,,x n } V assigns nodes to free variables 19

61 Binding To evaluate a formula φ we need a graph G=(V,E) and a binding α that maps free variables of φ to nodes of G. G,α satisfy φ φ is satisfiable G α φ(x 1,,x n ) α : {x 1,,x n } V assigns nodes to free variables The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) G α φ if α = {x v} v v' G v'' 19

62 Binding To evaluate a formula φ we need a graph G=(V,E) and a binding α that maps free variables of φ to nodes of G. G,α satisfy φ φ is satisfiable G α φ(x 1,,x n ) α : {x 1,,x n } V assigns nodes to free variables The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) G α φ if α = {x v} v v' Every node has at least two neighbours ψ = x y z ( (y=z) E(x,y) E(x,z)) G ψ G v'' 19

63 First-order logic Formal Semantics of FO G α x φ iff for some v V and α = α {x v} we have G α' φ G α x φ iff for every v V and α = α {x v} we have G α' φ G α φ ψ iff G α φ and G α ψ G α φ iff it is not true that G α φ G α x=y iff α(x)=α(y) G α E(x,y) iff (α(x),α(y)) E 20

64 Formulas as queries φ(x 1,, x n ) evaluated on G=(V,E) yields all the bindings that satisfy φ: φ(g) = { ( α(x 1 ),,α(x n ) ) G α φ, α: {x 1,,x n } V } 21

65 Formulas as queries φ(x 1,, x n ) evaluated on G=(V,E) yields all the bindings that satisfy φ: φ(g) = { ( α(x 1 ),,α(x n ) ) G α φ, α: {x 1,,x n } V } The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) Return all nodes with at least two neighbours 21

66 Formulas as queries φ(x 1,, x n ) evaluated on G=(V,E) yields all the bindings that satisfy φ: φ(g) = { ( α(x 1 ),,α(x n ) ) G α φ, α: {x 1,,x n } V } The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) φ(g) = {v, v', v''} φ(g') = {v, v'} Return all nodes with at least two neighbours v v' v v' v'' G v'' G' 21

67 Formulas as queries φ(x 1,, x n ) evaluated on G=(V,E) yields all the bindings that satisfy φ: φ(g) = { ( α(x 1 ),,α(x n ) ) G α φ, α: {x 1,,x n } V } The node x has at least two neighbours φ(x) = y z ( (y=z) E(x,y) E(x,z)) φ(g) = {v, v', v''} φ(g') = {v, v'} Return all nodes with at least two neighbours v v' v v' Every node has two neighbours ψ = x y z ( (y=z) E(x,y) E(x,z)) ψ(g) = {()} set with one element: the 0-tuple ψ(g') = {} empty set v'' G v'' G' 21

68 Question: Which bindings α verify G α φ for φ(x,y) = z (E(x,z) E(z,y)) v' and G = v? v'' 22

69 Question: Which bindings α verify G α φ for φ(x,y) = z (E(x,z) E(z,y)) v' and G = v? v'' Answer: α = { x v, y v }, α = { x v, y v }, φ(g) = {v,v', v''} {v,v', v''} α = { x v', y v }, and all the rest 22

70 Formulas as queries FO can serve as a declarative query language on relational databases : we express the properties of the answer Tables = Relations Queries = Formulas Rows = Tuples 23

71 Formulas as queries FO can serve as a declarative query language on relational databases : we express the properties of the answer Tables = Relations Queries = Formulas Rows = Tuples Particular to databases: Use of constants No functions Finite structure Quantification over active domain 23

72 Formulas as queries FO can serve as a declarative query language on relational databases : we express the properties of the answer Tables = Relations Queries = Formulas Rows = Tuples Particular to databases: Use of constants No functions Finite structure Quantification over active domain finite is different from 23

73 Formulas as queries FO can serve as a declarative query language on relational databases : we express the properties of the answer Tables = Relations Queries = Formulas Rows = Tuples Particular to databases: Use of constants No functions Finite structure Quantification over active domain finite is different from There are formulas φ that are satisfiable only on infinite structures. Like which? 23

74 Formulas as queries FO can serve as a declarative query language on relational databases : we express the properties of the answer Tables = Relations Queries = Formulas Rows = Tuples Particular to databases: Use of constants No functions Finite structure Quantification over active domain finite is different from There are formulas φ that are satisfiable only on infinite structures. Like which? φ = R(x,y) is an infinite linear order 23

75 Formulas as queries FO can serve as a declarative query language on relational databases : we express the properties of the answer Tables = Relations Queries = Formulas Rows = Tuples Particular to databases: Use of constants No functions Finite structure Quantification over active domain finite is different from There are formulas φ that are satisfiable only on infinite structures. Like which? φ = R(x,y) is an infinite linear order Finite model theory 23

76 Formulas as queries FO can serve as a declarative query language on relational databases : we express the properties of the answer Tables = Relations Queries = Formulas Rows = Tuples RA =* FO How = What RA and FO logic have roughly* the same expressive power! [E.F. Codd 1972] *FO without functions, with equality, on finite domains, 24

77 Formulas as queries RA FO R 1 R 2 R 1 (x 1,, x n ) R 2 (x n+1,, x m ) R 1 R 2 R 1 (x 1,, x n ) R 2 (x 1,, x n ) σ {i1 =j 1,,i n =j n } (R) R(x 1,, x m ) (x i1 =x j1 ) (x in =x jn ) π {i1,,i n } (R) ({x 1,,x m } \ {x i1,,x in }). R(x 1,, x m ) R 1 \ R 2 R 1 (x 1,, x n ) R 2 (x 1,, x n ) 25

78 Formulas as queries FO RA does not hold in general! 26

79 Formulas as queries FO RA does not hold in general! the complement of R RA FO : R(x) 26

80 Formulas as queries FO RA / the complement of R RA FO : R(x) 26

81 Formulas as queries FO RA / the complement of R RA FO : R(x) We restrict variables to range over active domain 26

82 Formulas as queries FO RA / the complement of R RA FO : R(x) We restrict variables to range over active domain elements in the relations FO act = FO restricted to active domain 26

83 Formulas as queries FO RA / the complement of R RA FO : R(x) We restrict variables to range over active domain elements in the relations FO act = FO restricted to active domain φ 1 (x,y) = E(x,y) φ 1 (G) = {(v 1,v 1 ),(v 3,v 1 )} φ 2 (x) = y E(y,x) φ 2 (G) = {v 2 } G = v 1 v 2 v 4 v 3 26

84 First-order logic restricted to active domain Formal Semantics of FO act G α x φ iff for some v ACT(G) and α = α {x v} we have G α' φ G α x φ iff for every v ACT(G) and α = α {x v} we have G α' φ G α φ ψ iff G α φ and G α ψ G α φ iff it is not true that G α φ G α x=y iff α(x)=α(y) G α E(x,y) iff (α(x),α(y)) E ACT(G) = {v for some v': (v,v') E or (v',v) E} 27

85 ` FO act RA 28

86 ` FO act RA Assume: 1. φ has variables x 1,,x n, 2. φ in normal form: ( * ( )*)* + quantifier-free ψ(x 1,,x n ) x 1,x 2 x 3 x 4. ( E(x 1,x 3 ) E(x 4,x 2 ) ) (x 1 =x 3 ) 28

87 ` FO act RA Assume: 1. φ has variables x 1,,x n, 2. φ in normal form: ( * ( )*)* + quantifier-free ψ(x 1,,x n ) x 1,x 2 x 3 x 4. ( E(x 1,x 3 ) E(x 4,x 2 ) ) (x 1 =x 3 ) Translation Adom = RA expression for active domain = π 1 (E) π 2 (E) ( x i φ(x i1,,x in ) ) π {i1,,i n }\{i} ( φ ) (x i = x j ) σ {i=j} ( Adom Adom ) (ψ 1 (x 1,,x n ) ψ 2 (x 1,,x n )) ψ 1 ψ 2 ( φ(x i1,,x in )) Adom Adom \ φ 28

88 ` FO act RA Assume: 1. φ has variables x 1,,x n, 2. φ in normal form: ( * ( )*)* + quantifier-free ψ(x 1,,x n ) x 1,x 2 x 3 x 4. ( E(x 1,x 3 ) E(x 4,x 2 ) ) (x 1 =x 3 ) Translation Adom = RA expression for active domain = π 1 (E) π 2 (E) ( x i φ(x i1,,x in ) ) π {i1,,i n }\{i} ( φ ) (x i = x j ) σ {i=j} ( Adom Adom ) A B = (A B) \ A \ B (ψ 1 (x 1,,x n ) ψ 2 (x 1,,x n )) ψ 1 ψ 2 ( φ(x i1,,x in )) Adom Adom \ φ 28

89 Corollary FO act is equivalent to RA 29

90 Question 1: How is π 2 (σ 1=3 (π 2 (σ 1=3 (R 1 R 2 )) R 2 )) expressed in FO? Remember: R 1,R 2 are binary Question 2: How is y,z. (R 1 (x,y) R 1 (y,z) x z ) expressed in RA? Remember: The signature is the same as before (R 1,R 2 binary) R 1 R 2 R 1 R 2 R 1 \ R 2 σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} π {i1,,i n } (R) {(x i 1,,x in ) (x 1,, x m ) R} 30

91 Question 1: How is π 2 (σ 1=3 (π 2 (σ 1=3 (R 1 R 2 )) R 2 )) expressed in FO? Remember: R 1,R 2 are binary Answer: x 2. ( x 1,x 4. (R 1 (x 1,x 2 ) R 2 (x 1,x 4 )) R 2 (x 2,x 5 ) ) Question 2: How is y,z. (R 1 (x,y) R 1 (y,z) x z ) expressed in RA? Remember: The signature is the same as before (R 1,R 2 binary) R 1 R 2 R 1 R 2 R 1 \ R 2 σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} π {i1,,i n } (R) {(x i 1,,x in ) (x 1,, x m ) R} 30

92 Question 1: How is π 2 (σ 1=3 (π 2 (σ 1=3 (R 1 R 2 )) R 2 )) expressed in FO? Remember: R 1,R 2 are binary Answer: x 2. ( x 1,x 4. (R 1 (x 1,x 2 ) R 2 (x 1,x 4 )) R 2 (x 2,x 5 ) ) Question 2: How is y,z. (R 1 (x,y) R 1 (y,z) x z ) expressed in RA? Remember: The signature is the same as before (R 1,R 2 binary) R 1 R 2 R 1 R 2 R 1 \ R 2 σ {i1 =j 1,,i n =j n } (R) {(x 1,, x m ) R (x i1 =x j1 ) (x in =x jn )} π {i1,,i n } (R) {(x i 1,,x in ) (x 1,, x m ) R} Answer: π 1 (σ {2=3,1 4} (R 1 R 1 )) 30

93 FO RA SQL = = Logic Algebra Programming language 31

94 over active domain on finite domains very basic FO RA SQL = = Logic Algebra Programming language 31

95 Algorithmic problems for query languages Evaluation problem: Given a query Q, a database instance db, and a tuple t, is t Q(db)? How hard is it to retrieve data? 32

96 Algorithmic problems for query languages Evaluation problem: Given a query Q, a database instance db, and a tuple t, is t Q(db)? How hard is it to retrieve data? Emptiness problem: Given a query Q, is there a database instance db so that Q(db)? Does Q make sense? Is it a contradiction? (Query optimization) 32

97 Algorithmic problems for query languages Evaluation problem: Given a query Q, a database instance db, and a tuple t, is t Q(db)? How hard is it to retrieve data? Emptiness problem: Given a query Q, is there a database instance db so that Q(db)? Does Q make sense? Is it a contradiction? (Query optimization) Equivalence problem: Given queries Q 1, Q 2, is Q 1 (db) = Q 2 (db) for all database instances db? Can we safely replace a query with another? (Query optimization) 32

98 Complexity theory What can be mechanized? decidable/undecidable How hard is it to mechanise? complexity classes 33

99 Complexity theory H s 10th Domino PCP What can be mechanized? decidable/undecidable How hard is it to mechanise? complexity classes K... 33

100 Complexity theory H s 10th Domino PCP What can be mechanized? decidable/undecidable How hard is it to mechanise? complexity classes K... usage of resources: time memory 33

101 Complexity theory H s 10th Domino PCP What can be mechanized? decidable/undecidable How hard is it to mechanise? complexity classes K... usage of resources: time memory Algorithm Alg is TIME-bounded by a function f : N N if Alg(input) uses less than f ( input ) units of TIME. 33

102 Complexity theory H s 10th Domino PCP What can be mechanized? decidable/undecidable How hard is it to mechanise? complexity classes K... usage of resources: time memory Algorithm Alg is TIME-bounded by a function f : N N if Alg(input) uses less than f ( input ) units of TIME. time input size f Alg 33

103 Complexity theory H s 10th Domino PCP What can be mechanized? decidable/undecidable How hard is it to mechanise? complexity classes K... usage of resources: time memory SPACE Algorithm Alg is TIME-bounded by a function f : N N if SPACE. Alg(input) uses less than f ( input ) units of TIME. time input size f Alg 33

104 Complexity theory H s 10th Domino PCP What can be mechanized? decidable/undecidable How hard is it to mechanise? complexity classes K... usage of resources: time memory SPACE Algorithm Alg is TIME-bounded by a function f : N N if SPACE. Alg(input) uses less than f ( input ) units of TIME. time input size f Alg LOGSPACE PTIME PSPACE EXPTIME 33

105 Complexity theory H s 10th Domino PCP What can be mechanized? decidable/undecidable How hard is it to mechanise? complexity classes K... usage of resources: time memory SPACE Algorithm Alg is TIME-bounded by a function f : N N if SPACE. Alg(input) uses less than f ( input ) units of TIME. time input size f Alg TIME-bounded by a polynomial LOGSPACE PTIME PSPACE EXPTIME SPACE-bounded by log(n) SPACE-bounded by a polynomial 33

106 Algorithmic problems for FO Evaluation problem: Given a FO formula φ(x 1,, x n ), a graph G, and a binding α, does G α φ? Satisfiability problem: Given a FO formula φ, is there a graph G and binding α, such that G α φ? Equivalence problem: Given FO formulae φ,ψ, is G α φ iff G α ψ for all graphs G and bindings α? 34

107 Algorithmic problems for FO Evaluation problem: Given a FO formula φ(x 1,, x n ), a graph G, and a binding α, does G α φ? DECIDABLE foundations of the database industry Satisfiability problem: Given a FO formula φ, is there a graph G and binding α, such that G α φ? Equivalence problem: Given FO formulae φ,ψ, is G α φ iff G α ψ for all graphs G and bindings α? 34

108 Algorithmic problems for FO Evaluation problem: Given a FO formula φ(x 1,, x n ), a graph G, and a binding α, does G α φ? DECIDABLE foundations of the database industry Satisfiability problem: Given a FO formula φ, is there a graph G and binding α, such that G α φ? UNDECIDABLE both for and finite Equivalence problem: Given FO formulae φ,ψ, is G α φ iff G α ψ for all graphs G and bindings α? 34

109 Algorithmic problems for FO Evaluation problem: Given a FO formula φ(x 1,, x n ), a graph G, and a binding α, does G α φ? DECIDABLE foundations of the database industry Satisfiability problem: Given a FO formula φ, is there a graph G and binding α, such that G α φ? UNDECIDABLE both for and finite Equivalence problem: Given FO formulae φ,ψ, is G α φ iff G α ψ for all graphs G and bindings α? UNDECIDABLE by reduction to the satisfiability problem 34

110 Algorithmic problems for FO Satisfiability problem: Given a FO formula φ, is there a graph G and binding α, such that G α φ? UNDECIDABLE both for and finite [Trakhtenbrot 50] 35

111 Algorithmic problems for FO Satisfiability problem: Given a FO formula φ, is there a graph G and binding α, such that G α φ? UNDECIDABLE both for and finite [Trakhtenbrot 50] Proof: By reduction from the Domino (aka Tiling) problem. 35

112 Algorithmic problems for FO Satisfiability problem: Given a FO formula φ, is there a graph G and binding α, such that G α φ? UNDECIDABLE both for and finite [Trakhtenbrot 50] Proof: By reduction from the Domino (aka Tiling) problem. Reduction from P to P': Algorithm that solves P using a O(1) procedure P'(x) that returns the truth value of P'(x). 35

113 The (undecidable) Domino problem Domino Input: 4-sided dominos:

114 The (undecidable) Domino problem Domino Input: 4-sided dominos: Output: Is it possible to form a white-bordered rectangle? (of any size)

115 The (undecidable) Domino problem Domino Input: 4-sided dominos: Output: Is it possible to form a white-bordered rectangle? (of any size) Rules: sides must match, you can t rotate the dominos, but you can clone them.

116 The (undecidable) Domino problem Domino - Why is it undecidable? It can easily encode halting computations of Turing machines: q 0 q q 0 0 l l l l 0 1 r r 0 0 r r 0 q q q 0 q 0 0 l l l l 1 r r r r s s

117 The (undecidable) Domino problem Domino - Why is it undecidable? It can easily encode halting computations of Turing machines: (head is elsewhere, symbol is not modified) q 0 q q 0 0 l l l l 0 1 r r 0 0 r r 0 q q q 0 q 0 0 l l l l 1 r r r r s s

118 The (undecidable) Domino problem Domino - Why is it undecidable? It can easily encode halting computations of Turing machines: r 2 r r q (head is elsewhere, symbol is not modified) 2 (head is here, symbol is rewritten, head moves right) 0 0 q 0 q q 0 0 l l l l 0 1 r r 0 0 r r 0 q q q 0 q 0 0 l l l l 1 r r r r s s

119 The (undecidable) Domino problem Domino - Why is it undecidable? It can easily encode halting computations of Turing machines: r 2 r r q 0 2 l 2 l 1 l l 2 q 0 2 (head is elsewhere, symbol is not modified) 2 (head is here, symbol is rewritten, head moves right) (head is here, symbol is rewritten, head moves left) 0 0 q 0 q q 0 0 l l l l 0 1 r r 0 0 r r 0 q q q 0 q 0 0 l l l l 1 r r r r s s

120 The (undecidable) Domino problem Domino - Why is it undecidable? It can easily encode halting computations of Turing machines: r 2 r r q 0 2 l 2 l 1 l l 2 q 0 2 (head is elsewhere, symbol is not modified) 2 s (head is here, symbol is rewritten, head moves right) (head is here, symbol is rewritten, head moves left) (initial configuration) 0 0 q 0 q q 0 0 l l l l 0 1 r r 0 0 r r 0 q q q 0 q 0 0 l l l l 1 r r r r s s

121 The (undecidable) Domino problem Domino - Why is it undecidable? It can easily encode halting computations of Turing machines: r 2 r r q 0 2 l 2 l 1 l l 2 q 0 2 (head is elsewhere, symbol is not modified) 2 s h (head is here, symbol is rewritten, head moves right) (head is here, symbol is rewritten, head moves left) (initial configuration) (halting configuration) 0 0 q 0 q q 0 0 l l l l 0 1 r r 0 0 r r 0 q q q 0 q 0 0 l l l l 1 r r r r s s

122 Domino Sat-FO (domino has a solution iff φ satisfiable) 1. There is a grid: H(, ) and V(, ) are relations representing bijections such that

123 Domino Sat-FO (domino has a solution iff φ satisfiable) 1. There is a grid: H(, ) and V(, ) are relations representing bijections such that V H

124 Domino Sat-FO (domino has a solution iff φ satisfiable) 1. There is a grid: H(, ) and V(, ) are relations representing bijections such that H V H V

125 Domino Sat-FO (domino has a solution iff φ satisfiable) 1. There is a grid: H(, ) and V(, ) are relations representing bijections such that H H H H V V V V V H V H V H H V... V V V V H H H H... V H V H V H... H V

126 Domino Sat-FO (domino has a solution iff φ satisfiable) 1. There is a grid: H(, ) and V(, ) are relations representing bijections such that H H H H V V V H H H V V V V V H V H V H H V... V V V V H H H H V H 2. Assign one domino to each node: a unary relation D ( x ) for each domino

127 Domino Sat-FO (domino has a solution iff φ satisfiable) 1. There is a grid: H(, ) and V(, ) are relations representing bijections such that H H H H V V V H H H V V V V V H V H V H H V... V V V V H H H H V H 2. Assign one domino to each node: a unary relation D ( x ) for each domino 3. Match the sides x,y if H(x,y), then D a (x) D b (y) for some dominos a,b that match horizontally (Idem vertically)

128 Domino Sat-FO (domino has a solution iff φ satisfiable) 1. There is a grid: H(, ) and V(, ) are relations representing bijections such that H H H H V V V V V V V H H H... V H V H V H H V... V V V V H H H H 4. Borders are white V H 2. Assign one domino to each node: a unary relation D ( x ) for each domino 3. Match the sides x,y if H(x,y), then D a (x) D b (y) for some dominos a,b that match horizontally (Idem vertically)

129 Algorithmic problems for FO Evaluation problem: Given a FO formula φ(x 1,, x n ), a graph G, and a binding α, does G α φ? DECIDABLE foundations of the database industry Satisfiability problem: Given a FO formula φ, is there a graph G and binding α, such that G α φ? UNDECIDABLE both for and finite Equivalence problem: Given FO formulae φ,ψ, is G α φ iff G α ψ for all graphs G and bindings α? UNDECIDABLE by reduction to the satisfiability problem 39

130 Algorithmic problems for FO Equivalence problem: Given FO formulae φ,ψ, is G α φ iff G α ψ for all graphs G and bindings α? UNDECIDABLE by reduction to the satisfiability problem 40

131 Algorithmic problems for FO φ is satisfiable iff φ is not equivalent to Satisfiability problem undecidable Equivalence problem undecidable Equivalence problem: Given FO formulae φ,ψ, is G α φ iff G α ψ for all graphs G and bindings α? UNDECIDABLE by reduction to the satisfiability problem 40

132 Algorithmic problems for FO φ is satisfiable iff φ is not equivalent to Satisfiability problem undecidable Equivalence problem undecidable Actually, there are reductions in both senses: φ(x 1,,x n ) and ψ(y 1,,y m ) are equivalent iff n=m (x 1 =y 1 ) (x n =y n ) φ(x 1,,x n ) ψ(y 1,,y n ) is unsatisfiable (x 1 =y 1 ) (x n =y n ) ψ(x 1,,x n ) φ(y 1,,y n ) is unsatisfiable Equivalence problem: Given FO formulae φ,ψ, is G α φ iff G α ψ for all graphs G and bindings α? UNDECIDABLE by reduction to the satisfiability problem 40

133 Algorithmic problems for FO Evaluation problem: Given a FO formula φ(x 1,, x n ), a graph G, and a binding α, does G α φ? DECIDABLE foundations of the database industry Satisfiability problem: Given a FO formula φ, is there a graph G and binding α, such that G α φ? UNDECIDABLE both for and finite Equivalence problem: Given FO formulae φ,ψ, is G α φ iff G α ψ for all graphs G and bindings α? UNDECIDABLE by reduction to the satisfiability problem 41

134 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V 42

135 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V Encoding of G = (V, E) each node is coded with a bit string of size log( V ), edge set is encoded by its tuples, e.g. (100,101), (010, 010), Cost of coding: G = E 2 log( V ) V (mod a polynomial) 42

136 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V Encoding of G = (V, E) each node is coded with a bit string of size log( V ), edge set is encoded by its tuples, e.g. (100,101), (010, 010), Cost of coding: G = E 2 log( V ) V (mod a polynomial) Encoding of α = {x 1,,x n } V each node is coded with a bit string of size log( V ), Cost of coding: α = n log( V ) 42

137 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V 43

138 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V If φ(x 1,,x n ) = E(x i,x j ): answer YES iff (α(x i ),α(x j )) E If φ(x 1,,x n ) = ψ(x 1,,x n ) ψ'(x 1,,x n ): answer YES iff G α ψ and G α ψ' If φ(x 1,,x n ) = ψ(x 1,,x n ): answer NO iff G α ψ If φ(x 1,,x n ) = y. ψ(x 1,,x n,y): answer YES iff for some v V and α'= α {y v} we have G α' ψ. 43

139 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V If φ(x 1,,x n ) = E(x i,x j ): answer YES iff (α(x i ),α(x j )) E If φ(x 1,,x n ) = ψ(x 1,,x n ) ψ'(x 1,,x n ): answer YES iff G α ψ and G α ψ' If φ(x 1,,x n ) = ψ(x 1,,x n ): answer NO iff G α ψ If φ(x 1,,x n ) = y. ψ(x 1,,x n,y): answer YES iff for some v V and α'= α {y v} we have G α' ψ. Question: How much space does it take? 43

140 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V If φ(x 1,,x n ) = E(x i,x j ): answer YES iff (α(x i ),α(x j )) E use 4 pointers LOGSPACE If φ(x 1,,x n ) = ψ(x 1,,x n ) ψ'(x 1,,x n ): answer YES iff G α ψ and G α ψ' If φ(x 1,,x n ) = ψ(x 1,,x n ): answer NO iff G α ψ If φ(x 1,,x n ) = y. ψ(x 1,,x n,y): answer YES iff for some v V and α'= α {y v} we have G α' ψ. Question: How much space does it take? 43

141 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V If φ(x 1,,x n ) = E(x i,x j ): answer YES iff (α(x i ),α(x j )) E If φ(x 1,,x n ) = ψ(x 1,,x n ) ψ'(x 1,,x n ): answer YES iff G α ψ and G α ψ' use 4 pointers LOGSPACE MAX( SPACE(G α ψ)), SPACE(G α ψ')) ) If φ(x 1,,x n ) = ψ(x 1,,x n ): answer NO iff G α ψ If φ(x 1,,x n ) = y. ψ(x 1,,x n,y): answer YES iff for some v V and α'= α {y v} we have G α' ψ. Question: How much space does it take? 43

142 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V If φ(x 1,,x n ) = E(x i,x j ): answer YES iff (α(x i ),α(x j )) E If φ(x 1,,x n ) = ψ(x 1,,x n ) ψ'(x 1,,x n ): answer YES iff G α ψ and G α ψ' If φ(x 1,,x n ) = ψ(x 1,,x n ): answer NO iff G α ψ use 4 pointers LOGSPACE MAX( SPACE(G α ψ)), SPACE(G α ψ')) ) SPACE(G α ψ)) If φ(x 1,,x n ) = y. ψ(x 1,,x n,y): answer YES iff for some v V and α'= α {y v} we have G α' ψ. Question: How much space does it take? 43

143 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V If φ(x 1,,x n ) = E(x i,x j ): answer YES iff (α(x i ),α(x j )) E If φ(x 1,,x n ) = ψ(x 1,,x n ) ψ'(x 1,,x n ): answer YES iff G α ψ and G α ψ' If φ(x 1,,x n ) = ψ(x 1,,x n ): answer NO iff G α ψ use 4 pointers LOGSPACE MAX( SPACE(G α ψ)), SPACE(G α ψ')) ) SPACE(G α ψ)) If φ(x 1,,x n ) = y. ψ(x 1,,x n,y): answer YES iff for some v V and α'= α {y v} we have G α' ψ. 2 log( G ) + SPACE(G α' ψ ) Question: How much space does it take? 43

144 Evaluation problem for FO φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V If φ(x 1,,x n ) = E(x i,x j ): answer YES iff (α(x i ),α(x j )) E If φ(x 1,,x n ) = ψ(x 1,,x n ) ψ'(x 1,,x n ): answer YES iff G α ψ and G α ψ' If φ(x 1,,x n ) = ψ(x 1,,x n ): answer NO iff G α ψ use 4 pointers LOGSPACE MAX( SPACE(G α ψ)), SPACE(G α ψ')) ) SPACE(G α ψ)) If φ(x 1,,x n ) = y. ψ(x 1,,x n,y): answer YES iff for some v V and α'= α {y v} we have G α' ψ. 2 log( G ) + SPACE(G α' ψ ) Question: 2 log( G ) log( G ) + k log( α + G ) space How much space does it take? φ times 43

145 Evaluation problem for FO in PSPACE φ(x 1,,x n ) Input: G = (V,E) Output: G α φ? α = {x 1,,x n } V If φ(x 1,,x n ) = E(x i,x j ): answer YES iff (α(x i ),α(x j )) E If φ(x 1,,x n ) = ψ(x 1,,x n ) ψ'(x 1,,x n ): answer YES iff G α ψ and G α ψ' If φ(x 1,,x n ) = ψ(x 1,,x n ): answer NO iff G α ψ use 4 pointers LOGSPACE MAX( SPACE(G α ψ)), SPACE(G α ψ')) ) SPACE(G α ψ)) If φ(x 1,,x n ) = y. ψ(x 1,,x n,y): answer YES iff for some v V and α'= α {y v} we have G α' ψ. 2 log( G ) + SPACE(G α' ψ ) Question: 2 log( G ) log( G ) + k log( α + G ) space How much space does it take? φ times 43

146 Combined, Query, and Data complexities [Vardi, 1982] Problem: Usual scenario in database A database of size 10 6 A query of size

147 Combined, Query, and Data complexities [Vardi, 1982] Problem: Usual scenario in database A database of size 10 6 A query of size 100 Input: database + query 44

148 Combined, Query, and Data complexities [Vardi, 1982] Problem: Usual scenario in database A database of size 10 6 A query of size 100 Input: database + query TIME(2 query + data ) But we don t distinguish this in the analysis: = TIME( query + 2 data ) 44

149 Combined, Query, and Data complexities [Vardi, 1982] Query and data play very different roles. Separation of concerns: How the resources grow with respect to the size of the data the query size 45

150 Combined, Query, and Data complexities Combined complexity: input size is query + data Query complexity ( data fixed): input size is query Data complexity ( query fixed): input size is data 46

151 Combined, Query, and Data complexities Combined complexity: input size is query + data Query complexity ( data fixed): input size is query Data complexity ( query fixed): input size is data O(2 query + data ) is O( query + 2 data ) is exponential in combined complexity exponential in query complexity linear in data complexity exponential in combined complexity linear in query complexity exponential in data complexity 46

152 Question What is the data, query and combined complexity for the evaluation problem for FO? Remember: data complexity, input size: data query complexity, input size: query combined complexity, input size: data + query φ 2 log( G ) + k log( α + G ) space 47

153 Question What is the data, query and combined complexity for the evaluation problem for FO? Remember: data complexity, input size: data query complexity, input size: query combined complexity, input size: data + query φ 2 log( G ) + k log( α + G ) space query data O(log( data ) query ) space PSPACE combined and query complexity LOGSPACE data complexity 47

154 Evaluation pb for FO is PSPACE-complete (combined complexity) PSPACE-complete problem: QBF (satisfaction of Quantified Boolean Formulas) QBF = a boolean formula with quantification over the truth values (T,F) 48

155 Evaluation pb for FO is PSPACE-complete (combined complexity) PSPACE-complete problem: QBF (satisfaction of Quantified Boolean Formulas) QBF = a boolean formula with quantification over the truth values (T,F) p q. (p q) where p,q range over {T,F} 48

156 Evaluation pb for FO is PSPACE-complete (combined complexity) PSPACE-complete problem: QBF (satisfaction of Quantified Boolean Formulas) QBF = a boolean formula with quantification over the truth values (T,F) p q. (p q) where p,q range over {T,F} Theorem: Evaluation for FO is PSPACE-complete (combined c.) 48

157 Evaluation pb for FO is PSPACE-complete (combined complexity) PSPACE-complete problem: QBF (satisfaction of Quantified Boolean Formulas) QBF = a boolean formula with quantification over the truth values (T,F) p q. (p q) where p,q range over {T,F} Theorem: Evaluation for FO is PSPACE-complete (combined c.) Polynomial reduction QBF FO : 1. Given ψ QBF, let ψ (x) be the replacement of each p with p=x in ψ. 2. Note: x ψ' holds in a 2-element graph iff ψ is QBF-satisfiable 3. Test if G ψ' for G=({v,v'},{}) 48

158 Evaluation pb for FO is PSPACE-complete (combined complexity) PSPACE-complete problem: QBF (satisfaction of Quantified Boolean Formulas) QBF = a boolean formula with quantification over the truth values (T,F) p q. (p q) where p,q range over {T,F} Theorem: Evaluation for FO is PSPACE-complete (combined c.) Polynomial reduction QBF FO : ψ'(x)= p q. ( (p=x) (q=x) ) 1. Given ψ QBF, let ψ (x) be the replacement of each p with p=x in ψ. 2. Note: x ψ' holds in a 2-element graph iff ψ is QBF-satisfiable 3. Test if G ψ' for G=({v,v'},{}) 48

159 Evaluation pb for FO is PSPACE-complete (combined complexity) PSPACE-complete problem: QBF (satisfaction of Quantified Boolean Formulas) QBF = a boolean formula with quantification over the truth values (T,F) p q. (p q) where p,q range over {T,F} Theorem: Evaluation for FO is PSPACE-complete (combined c.) Polynomial reduction QBF FO : ψ'(x)= p q. ( (p=x) (q=x) ) x p q. ( (p=x) (q=x) ) 1. Given ψ QBF, let ψ (x) be the replacement of each p with p=x in ψ. 2. Note: x ψ' holds in a 2-element graph iff ψ is QBF-satisfiable 3. Test if G ψ' for G=({v,v'},{}) 48

160 Recap Equivalence-RA Equivalence-SQL Domino Equivalence-FO Sat-FO Eval-FO (combined) QBF Eval-FO (data) UNDECIDABLE PSPACE LOGSPACE 49

161 Bibliography Abiteboul, Hull, Vianu, Foundations of Databases, Addison-Wesley, (freely available at Chapters 1, 2, 3 50