Lossless Horizontal Decomposition with Domain Constraints on Interpreted Attributes

Transcription

1 Lossless Horizontal Decomposition with Domain Constraints on Interpreted Attributes Ingo Feinerer 1 and Enrico Franconi 2 and Paolo Guagliardo 2 1 Vienna University of Technology 2 Free University of Bozen-Bolzano 9 th July 2013 BNCOD 2013, Oxford (UK)

2 Horizontal Decomposition R 1 / 19

3 Horizontal Decomposition V 1 selection R 1 / 19

4 Horizontal Decomposition V 2 V 1 selection selection R 1 / 19

5 Horizontal Decomposition V 2 V 1 V 3 selection selection selection R 1 / 19

6 Horizontal Decomposition V 2 V 1 V 3 R 1 / 19

7 Horizontal Decomposition V 2 V 1 V 3? R 1 / 19

8 Interpreted Attributes Data values from special domains (e.g., the integers or the reals) on which a set of predicates (e.g., smaller/greater than) and functions (e.g., addition and subtraction) are defined, according to a first-order theory C 2 / 19

9 Interpreted Attributes Data values from special domains (e.g., the integers or the reals) on which a set of predicates (e.g., smaller/greater than) and functions (e.g., addition and subtraction) are defined, according to a first-order theory C Notation R( x 1,..., x k, y 1,..., y n ) x 1,..., x k non-interpreted (values from dom) y 1,..., y n interpreted over idom associated with C 2 / 19

10 Interpreted Attributes Data values from special domains (e.g., the integers or the reals) on which a set of predicates (e.g., smaller/greater than) and functions (e.g., addition and subtraction) are defined, according to a first-order theory C Notation R( x 1,..., x k, y 1,..., y n ) x 1,..., x k non-interpreted (values from dom) y 1,..., y n interpreted over idom associated with C We assume C to be closed under negation 2 / 19

11 Conditional Domain Constraints (CDC) A conditional domain constraint is a formula of the form x, y. R(x, y) λ(x) δ(y) where λ(x) conjunction of (possibly negated) equalities x = a, with x from x and a from dom δ(y) a formula in C For short R: λ(x) δ(y) 3 / 19

12 View Definitions (Selections) Each view symbol V is defined by a formula of the form x, y. V (x, y) ( R(x, y) λ(x) σ(y) ) where λ(x) conjunction of (possibly negated) equalities x = a, with x from x and a from dom δ(y) a formula in C For short V : λ(x) σ(y) 4 / 19

13 Unit Two-Variable Per Inequality (UTVPI) Constraints A UTVPI constraint is a formula of the form ax + by d where x, y integer variables a, b { 1, 0, 1} (unit coefficients) d Z (an integer) 5 / 19

14 Unit Two-Variable Per Inequality (UTVPI) Constraints A UTVPI constraint is a formula of the form ax + by d where x, y integer variables a, b { 1, 0, 1} (unit coefficients) d Z (an integer) Fragment of linear arithmetic constraints over the integers Satisfiability of UTVPIs is decidable in polynomial time 5 / 19

15 Unit Two-Variable Per Inequality (UTVPI) Constraints A UTVPI constraint is a formula of the form ax + by d where x, y integer variables a, b { 1, 0, 1} (unit coefficients) d Z (an integer) Fragment of linear arithmetic constraints over the integers Satisfiability of UTVPIs is decidable in polynomial time butvpi = a Boolean combination of UTVPIs Satisfiability of butvpis is NP-complete 5 / 19

16 Running Example (1) Employee( x 1, x 2, x 3, y 1, y 2 ) Name Department Position Salary Bonus non-interpreted interpreted over Z 6 / 19

17 Running Example (1) Employee( x 1, x 2, x 3, y 1, y 2 ) Name Department Position Salary Bonus non-interpreted interpreted over Z Let y 1, y 2 be expressed in thousands of euros per month Let a = ICT and b = Manager 6 / 19

18 Running Example (2) Integrity constraints x 2 = a y 1 + y 2 5 x 3 = b y 2 2 y 1 y 2 0 Employees in ICT gain at most 5 Managers receive a bonus of at least 2 Bonus is never greater than salary 7 / 19

19 Running Example (2) Integrity constraints x 2 = a y 1 + y 2 5 x 3 = b y 2 2 y 1 y 2 0 Employees in ICT gain at most 5 Managers receive a bonus of at least 2 Bonus is never greater than salary View definitions Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b Managers in departments other than ICT Employees getting strictly less than 4 as bonus Employees not working as managers 7 / 19

20 Outline Losslessness General method (independent of C) for checking losslessness under CDCs Separability Capture interaction between CDCs and other types of constraints w.r.t. losslessness 8 / 19

21 Checking for Losslessness: Overview 1. Build propositional theories Π and Π Σ associated with and Σ, respectively 2. Build an auxiliary theory Π for Π = Π Π Σ 3. For every valuation α of the propositional variables in Π consider the α-filtering Π α consisting of consequents of the C-CDCs in that are applicable under α negated C-formulae appearing in selections of Σ whose equalities and inequalities are satisfied by α 9 / 19

22 Checking for Losslessness: Overview 1. Build propositional theories Π and Π Σ associated with and Σ, respectively 2. Build an auxiliary theory Π for Π = Π Π Σ 3. For every valuation α of the propositional variables in Π consider the α-filtering Π α consisting of consequents of the C-CDCs in that are applicable under α negated C-formulae appearing in selections of Σ whose equalities and inequalities are satisfied by α Theorem A horizontal decomposition Σ is lossless under if and only if the α-filtering Π α = Π α Πα Σ of Π is unsatisfiable for every valuation α of pvar(π) satisfying Π 9 / 19

23 Propositional Theories For ϕ Σ, let where prop(ϕ) = P v P propositional formula (possibly ) obtained from λ(x) in the antecedent of ϕ by replacing each x i = a with p a i v fresh propositional variable (possibly ) associated with the C-formula δ(y) or σ(y) appearing in ϕ 10 / 19

24 Propositional Theories For ϕ Σ, let where prop(ϕ) = P v P propositional formula (possibly ) obtained from λ(x) in the antecedent of ϕ by replacing each x i = a with p a i v fresh propositional variable (possibly ) associated with the C-formula δ(y) or σ(y) appearing in ϕ Propositional theories associated with and Σ: Π = { prop(ϕ) ϕ } Π Σ = { prop(ϕ) ϕ Σ } 10 / 19

25 Running Example Building the Propositional Theories Π and Π Σ x 2 = a y 1 + y 2 5 x 3 = b y 2 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b 11 / 19

26 Running Example Building the Propositional Theories Π and Π Σ x 2 = a y 1 + y 2 5 x 3 = b y 2 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b 11 / 19

27 Running Example Building the Propositional Theories Π and Π Σ p a 2 y 1 + y 2 5 x 3 = b y 2 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b 11 / 19

28 Running Example Building the Propositional Theories Π and Π Σ p a 2 y 1 + y 2 5 x 3 = b y 2 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b 11 / 19

29 Running Example Building the Propositional Theories Π and Π Σ p a 2 v 1 x 3 = b y 2 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y / 19

30 Running Example Building the Propositional Theories Π and Π Σ p a 2 v 1 x 3 = b y 2 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y / 19

31 Running Example Building the Propositional Theories Π and Π Σ p a 2 v 1 p b 3 y 2 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y / 19

32 Running Example Building the Propositional Theories Π and Π Σ p a 2 v 1 p b 3 y 2 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y / 19

33 Running Example Building the Propositional Theories Π and Π Σ p a 2 v 1 p b 3 v 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y 2 5 v 2 y / 19

34 Running Example Building the Propositional Theories Π and Π Σ p a 2 v 1 p b 3 v 2 y 1 y 2 0 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y 2 5 v 2 y / 19

35 Running Example Building the Propositional Theories Π and Π Σ p a 2 v 1 p b 3 v 2 v 3 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y 2 5 v 2 y 2 2 v 3 y 1 y / 19

36 Running Example Building the Propositional Theories Π and Π Σ Π p a 2 v 1 p b 3 v 2 v 3 Σ V 1 : x 2 a x 3 = b V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y 2 5 v 2 y 2 2 v 3 y 1 y / 19

37 Running Example Building the Propositional Theories Π and Π Σ Π p a 2 v 1 p b 3 v 2 v 3 Σ V 1 : p a 2 pb 3 V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y 2 5 v 2 y 2 2 v 3 y 1 y / 19

38 Running Example Building the Propositional Theories Π and Π Σ Π p a 2 v 1 p b 3 v 2 v 3 Σ V 1 : p a 2 pb 3 V 2 : y 2 < 4 V 3 : x 3 b constr = v 1 y 1 + y 2 5 v 2 y 2 2 v 3 y 1 y / 19

39 Running Example Building the Propositional Theories Π and Π Σ Π p a 2 v 1 p b 3 v 2 v 3 Σ V 1 : p a 2 pb 3 V 2 : v 4 V 3 : x 3 b constr = v 1 y 1 + y 2 5 v 2 y 2 2 v 3 y 1 y 2 0 v 4 y 2 > 4 11 / 19

40 Running Example Building the Propositional Theories Π and Π Σ Π p a 2 v 1 p b 3 v 2 v 3 Σ V 1 : p a 2 pb 3 V 2 : v 4 V 3 : x 3 b constr = v 1 y 1 + y 2 5 v 2 y 2 2 v 3 y 1 y 2 0 v 4 y 2 > 4 11 / 19

41 Running Example Building the Propositional Theories Π and Π Σ Π p a 2 v 1 p b 3 v 2 v 3 Σ V 1 : p a 2 pb 3 V 2 : v 4 V 3 : p b 3 constr = v 1 y 1 + y 2 5 v 2 y 2 2 v 3 y 1 y 2 0 v 4 y 2 > 4 11 / 19

42 Running Example Building the Propositional Theories Π and Π Σ Π p a 2 v 1 p b 3 v 2 v 3 Π Σ V 1 : p a 2 pb 3 V 2 : v 4 V 3 : p b 3 constr = v 1 y 1 + y 2 5 v 2 y 2 2 v 3 y 1 y 2 0 v 4 y 2 > 4 11 / 19

43 Running Example Building the Propositional Theories Π and Π Σ Π p a 2 v 1 p b 3 v 2 v 3 Π Σ V 1 : p a 2 pb 3 V 2 : v 4 V 3 : p b 3 constr = v 1 y 1 + y 2 5 v 2 y 2 2 v 3 y 1 y 2 0 v 4 y 2 > 4 Π = Π Π Σ ; pvar(π) = { p a 2, pb 3 } ; cvar(π) = { v 1, v 2, v 3, v 4 } 11 / 19

44 Auxiliary Theory no two distinct values in the same position Π = { p a i pb i a b, pa i, pb i pvar(π)} { (P ) Π Σ } 12 / 19

45 Auxiliary Theory no two distinct values in the same position Π = { p a i pb i a b, pa i, pb i pvar(π)} { (P ) Π Σ } In our running example: Π = { p a 2 pb 3, pb 3 } 12 / 19

46 Filtering α-filtering For a valuation α of pvar(π) Π α = { constr(v) (P v) Π, α(p ) = T } Π α Σ = { constr(v) (P v) Π Σ, α(p ) = T, v } 13 / 19

47 Filtering α-filtering For a valuation α of pvar(π) Π α = { constr(v) (P v) Π, α(p ) = T } Π α Σ = { constr(v) (P v) Π Σ, α(p ) = T, v } Theorem A horizontal decomposition Σ is lossless under if and only if the α-filtering Π α = Π α Πα Σ of Π is unsatisfiable for every valuation α of pvar(π) satisfying Π 13 / 19

48 Filtering α-filtering For a valuation α of pvar(π) Π α = { constr(v) (P v) Π, α(p ) = T } Π α Σ = { constr(v) (P v) Π Σ, α(p ) = T, v } Theorem A horizontal decomposition Σ is lossless under if and only if the α-filtering Π α = Π α Πα Σ of Π is unsatisfiable for every valuation α of pvar(π) satisfying Π Running Example The only valuation satisfying Π = { p a 2 pb 3, pb 3 } is α = { p b 3 T, pa 2 T } 13 / 19

49 Running Example: Checking for Losslessness α = { p b 3 T, pa 2 T } Π p a 2 v 1 p b 3 v 2 Π α = { constr(v 1), constr(v 2 ), constr(v 3 ) } v 3 Π Σ p a 2 pb 3 v 4 p b 3 Π α Σ = { constr(v 4) } 14 / 19

50 Running Example: Checking for Losslessness α = { p b 3 T, pa 2 T } Π p a 2 v 1 p b 3 v 2 v 3 Π Σ p a 2 pb 3 v 4 Π α = { constr(v 1), constr(v 2 ), constr(v 3 ) } y 1 + y 2 5 y 2 2 y 1 y 2 0 y 2 4 Π α Σ = { constr(v 4) } p b 3 14 / 19

51 Running Example: Checking for Losslessness α = { p b 3 T, pa 2 T } Π p a 2 v 1 p b 3 v 2 v 3 Π Σ p a 2 pb 3 v 4 Π α = { constr(v 1), constr(v 2 ), constr(v 3 ) } y 1 + y 2 5 y 2 2 y 1 y 2 0 y 2 4 Π α Σ = { constr(v 4) } p b 3 14 / 19

52 Running Example: Checking for Losslessness α = { p b 3 T, pa 2 T } Π p a 2 v 1 p b 3 v 2 v 3 Π Σ p a 2 pb 3 v 4 p b 3 Π α = { constr(v 1), constr(v 2 ), constr(v 3 ) } y 1 3 y 1 y 2 0 y 2 4 Π α Σ = { constr(v 4) } 14 / 19

53 Running Example: Checking for Losslessness α = { p b 3 T, pa 2 T } Π p a 2 v 1 p b 3 v 2 v 3 Π Σ p a 2 pb 3 v 4 p b 3 Π α = { constr(v 1), constr(v 2 ), constr(v 3 ) } y 1 3 y 1 + y 2 0 y 2 4 Π α Σ = { constr(v 4) } 14 / 19

54 Running Example: Checking for Losslessness α = { p b 3 T, pa 2 T } Π p a 2 v 1 p b 3 v 2 v 3 Π Σ p a 2 pb 3 v 4 p b 3 Π α = { constr(v 1), constr(v 2 ), constr(v 3 ) } y 1 3 y 2 3 y 2 4 Π α Σ = { constr(v 4) } 14 / 19

55 Additional Integrity Constraints: FDs Does allowing other types of integrity constraints along with CDCs make a difference w.r.t. losslessness? 15 / 19

56 Additional Integrity Constraints: FDs Does allowing other types of integrity constraints along with CDCs make a difference w.r.t. losslessness? For functional dependencies (FDs) the answer is NO Theorem Let consist of FDs and CDCs, and let consist of all the CDCs in. Then, a horizontal decomposition is lossless under if and only if it is lossless under. 15 / 19

57 Additional Integrity Constraints: FDs Does allowing other types of integrity constraints along with CDCs make a difference w.r.t. losslessness? For functional dependencies (FDs) the answer is NO Theorem Let consist of FDs and CDCs, and let consist of all the CDCs in. Then, a horizontal decomposition is lossless under if and only if it is lossless under. A horizontal decomposition is lossy under CDCs iff there is a one-tuple counter-example to its losslessness, and an FD violation always involves at least two tuples 15 / 19

58 Additional Integrity Constraints: UINDs What about unary inclusion dependencies (UINDs)? 16 / 19

59 Additional Integrity Constraints: UINDs What about unary inclusion dependencies (UINDs)? Example R(y 1, y 2 ) with y 1, y 2 interpreted over Z = { y 2 > 4, R[1] R[2] } Σ = {V 1 : y 1 > 3} 16 / 19

60 Additional Integrity Constraints: UINDs What about unary inclusion dependencies (UINDs)? Example R(y 1, y 2 ) with y 1, y 2 interpreted over Z = { y 2 > 4, R[1] R[2] } Σ = {V 1 : y 1 > 3} Does V 1 select all of the tuples in R? 16 / 19

61 Additional Integrity Constraints: UINDs What about unary inclusion dependencies (UINDs)? Example R(y 1, y 2 ) with y 1, y 2 interpreted over Z = { y 2 > 4, R[1] R[2] } Σ = {V 1 : y 1 > 3} Does V 1 select all of the tuples in R? YES! because = y 1 > 4 There is some interaction between CDCs and UINDs that influences losslessness 16 / 19

62 Separability C a class of integrity constraints S a finite set of sound inference rules for C extended with CDCs a set of CDCs and C-constraints 17 / 19

63 Separability C a class of integrity constraints S a finite set of sound inference rules for C extended with CDCs a set of CDCs and C-constraints Definition The C-constraints are S-separable in from the CDCs iff every horizontal decomposition is lossless under precisely when it is lossless under = {φ φ is a CDC}, where is the closure of w.r.t. S 17 / 19

64 Separability C a class of integrity constraints S a finite set of sound inference rules for C extended with CDCs a set of CDCs and C-constraints Definition The C-constraints are S-separable in from the CDCs iff every horizontal decomposition is lossless under precisely when it is lossless under = {φ φ is a CDC}, where is the closure of w.r.t. S S fully captures the interaction between C-constraints and CDCs w.r.t. losslessness losslessness can be checked by making explicit such interaction and then disregarding all constraints other than CDCs 17 / 19

65 Domain Propagation Can we fully capture the interaction between CDCs and UINDs? 18 / 19

66 Domain Propagation Can we fully capture the interaction between CDCs and UINDs? Domain Propagation Rule δ(y i ) R[y j ] R[y i ] δ(y j ) (dp) 18 / 19

67 Domain Propagation Can we fully capture the interaction between CDCs and UINDs? Domain Propagation Rule δ(y i ) R[y j ] R[y i ] δ(y j ) (dp) Theorem Let be a set of UINDs on interpreted positions and CDCs such that, for every UIND R[y i ] R[y j ] in, all of the CDCs in that mention y, where y is y i or y j, are of the form δ(y) Then, the UINDs are {(dp)}-separable in from the CDCs 18 / 19

68 Future Work Separability of FDs and UINDs (together) FDs and UINDs interact between them Interaction with other constraints (e.g., order dependencies) Turn a lossy decomposition into a lossless one find minimal fix: select exactly the missing tuples Update a fragment independently of the others 19 / 19