Foundations of Databases

Transcription

1 Foundations of Databases Free Universit of Bozen Bolzano, 2009 Werner Nutt (Most slides adapted from Thomas Eiter and Leonid Libkin) Foundations of Databases 1 Quer optimization: finding a good wa to evaluate a quer Queries are declarative, and can be translated into procedural languages in more than one wa Hence one has to choose the best (or at least good) procedural quer This happens in the contet of quer processing A quer processor turns queries and updates into sequences of operations on the database

2 Foundations of Databases 2 Stages Queries are translated into an etended relational algebra (operator + eecution method): Which algebra epressions will allow for an efficient eecution? Algebraic operators can be implemented b different methods (Eamples?): Which algorithm should be used for each operator? How do operators pass data (write into main memor, write on disk, pipeline into other operators)? Issues: Translate the quer into etended relational algebra ( quer plans ) Esimate the eecution of the plans: We need to know how data is stored, how it accessed, how large are intermediate results, etc. Decisions are based on general guidelines and statistical information Foundations of Databases 3 Overview of Quer Processing Start with a declarative quer: SELECT R.A, S.B, T.E FROM R,S,T WHERE R.C=S.C AND S.D=T.D AND R.A>5 AND S.B<3 AND T.D=T.E Translate into an algebra epression: π R.A,S.B,T.E (σ R.A>5 S.B<3 T.D=T.E (R S T)) Optimization step: rewrite to an equivalent but more efficient epression: π R.A,S.B,T.E (σ A>5 (R) σ B<3 (S) σ D=E (T))) Wh ma this be more efficient? Is there a still more efficient plan?

3 Foundations of Databases 4 Overview of Quer Processing (contd) When evaluating an epression with pushed selections, the main choice to make is the order of joins (ma influence selections). First quer plan (out of two): A,B A>5 B<3 D=E R S T first join S and T, and then join the result with R. Foundations of Databases 5 Overview of quer processing (contd) Alternative quer plan: A,B A>5 B<3 D=E R S T First join R and S, then join the result with T. Both quer plans produce the same result (wh?).

4 Foundations of Databases 6 Wh choose one and not the other? Foundations of Databases 7 Optimization b Algebraic Equivalences Given a relational algebra epression E, find another epression E equivalent to E that is easier (faster) to evaluate. Basic question: Given two relational algebra epressions E 1, E 2, are the equivalent? If there were a method to decide equivalence, we could turn it into one to decide whether an epression E is empt, i.e., whether E(I) = for ever instance I. How does one show this? Problem: Testing whether E is undecidable for epressions of full relational algebra i.e., including union and difference. (An idea wh?) Good news: We can still list some useful equalities. Equivalence is decidable for important classes of queries (SPJR, SPC)

5 Foundations of Databases 8 Optimization b Algebraic Equivalences (cntd) Sstematic wa of quer optimization: Appl equivalences and are commutative and associative, hence applicable in an order Cascaded projections can be simplified: If the attributes A 1,...,A n are among B 1,...,B m, then Cascaded selections might be merged: π A1,...,A n (π B1,...,B m (E)) = π A1,...,A n (E) σ C1 (σ C2 (E)) = σ C1 C 2 (E) Commuting selection with join. If c onl involves attributes from E 1, then etc We will not treat this here. σ C (E 1 E 2 ) = σ C (E 1 ) E 2 Foundations of Databases 9 Optimization b Algebraic Equivalences (contd) Rules combining σ,π with and \ Commuting selection and union: σ C (E 1 E 2 ) = σ C (E 1 ) σ C (E 2 ) Commuting selection and difference: σ C (E 1 \ E 2 ) = σ C (E 1 ) \ σ C (E 2 ) Commuting projection and union: π A1,...,A n (E 1 E 2 ) = π A1,...,A n (E 1 ) π A1,...,A n (E 2 ) Question: what about projection and difference? Is π A (E 1 \ E 2 ) equal to π A (E 1 ) \ π A (E 2 )?

6 Foundations of Databases 10 Optimization of Conjunctive Queries Reminder: Conjunctive queries = SPJR queries = simple SELECT-FROM-WHERE SQL queries (onl AND and (in)equalit in the WHERE clause) Etremel common, and thus special optimization techniques have been developed Reminder: for two relational algebra epressions E 1,E 2, E 1 = E 2 is undecidable. But for conjunctive queries, even E 1 E 2 is decidable. Main goal of optimizing conjunctive queries: reduce the number of joins. Foundations of Databases 11 Optimization of Conjunctive Queries: An Eample Given a relation R with two attributes A,B Two SQL queries: Q1 Q2 SELECT DISTINCT R1.B, R1.A SELECT DISTINCT R3.A, R1.A FROM R R1, R R2 FROM R R1, R R2, R R3 WHERE R2.A=R1.B WHERE R1.B=R2.B AND R2.B=R3.A Are the equivalent? If the are, we can save one join operation. In relational algebra: Q 1 = π 2,1 (σ 2=3 (R R)) Q 2 = π 5,1 (σ 2=4 4=5 (R R R))

7 Foundations of Databases 12 Optimization of Conjunctive Queries (contd) We will show that Q 1 and Q 2 are equivalent! We cannot do this b using our equivalences for relational algebra epression. (Wh?) Alternative idea: CQs are like patterns that have to be matched b a database. If a CQ returns an answer over a db, a part of the db must look like the quer. Use rule based notation to find a representation of part of the db: Q 1 (,) : Q 2 (,) : R(,),R(,z) R(,),R(w,),R(,u) Foundations of Databases 13 Containment is a Ke Propert Definition. (Quer containment) A quer Q is contained in a quer Q (written Q Q ) if Q(I) Q (I) for ever database instance I. Translations: If we can decide containment for a class of queries, then we can also decide equivalence. If Q is a class of queries that is closed under intersection, then the containment problem for Q can be reduced to the equivalence problem for Q.

8 Foundations of Databases 14 Checking Containment of Conjunctive Queries: Ideas We want to check containment of two CQs Q, Q We consider CQs without equalities and inequalities (How serious is this restriction?) We also assume that Q and Q have the same arit and identical vectors of head variables, that is, the are defined as Q( ) : B, Q ( ) : B Intuition: a conjunctive quer Q( ) : B returns an answer over instance I if I matches the pattern B Instead of checking containment over all instances, we consider onl finitel man test instances (or better, one!) Foundations of Databases 15 Tableau Notation of Conjunctive Queries Tableau notation blurs the distinction between quer and database instance We first consider queries over a single relation Q 1 (, ) : R(, ), R(, z) Q 2 (, ) : R(, ), R(w, ), R(, u) Tableau: A B A B z w answer line u Variables in the answer line are called distinguished answer line

9 Foundations of Databases 16 Tableau Homomorphisms Tableau (as well as database instances or first order structures) are compared b homomorphisms Idea: T is more general than T if there is a homomorphism from T to T Reminder: Terms are variables or constants A homomorphism δ from Tableau T 1 to tableau T 2 is a mapping such that δ: {variables of T 1 } {terms of T 2 } δ() = for ever distinguished if (t 1,...,t k ) is a row in T 1, then (δ(t 1 ),...,δ(t k )) is a row in T 2, where δ is etended to constants such that δ(a) = a for ever constant a Foundations of Databases 17 Tableau for Queries with Multiple Relations So far we assumed that there is onl one relation R, but what if there are man? Construct a tableau for the quer: Q(,): S(,),R(,z),R(,w),R(w, ) We create rows for each relation: A B S: A B R: z w w Formall, a tableau is just a database where variables can appear in tuples, plus a set of distinguished variables.

10 Foundations of Databases 18 Tableau Homomorphisms: General Case Let T 1, T 2 be tableau with the same distinguished variables A homomorphism δ from T 1 to T 2 is a mapping δ: {variables of T 1 } {terms of T 2 } such that δ(a) = a for ever constant δ() = for ever distinguished variable if t is a row of R in T 1, then δ( t) is a row of R in T 2, for ever relation R Foundations of Databases 19 The Homomorphism Theorem for Tableau Homomorphism Theorem: Let Q, Q be two conjunctive queries without equalities and inequalities that have the same distinguished variables and let T, T be their tableau. Then Q Q there eists a homomorphism from T to T

11 Foundations of Databases 20 Proof: Necessit Assume that Q Q Note that T, ignoring the declaration of distinguished variables, can be seen as a database instance (if we view variables as constants) Note also that Q(T) Since Q Q, it follows that Q (T) Since Q (T), there is a valuation ν such ν satisfies Q over T ν ( ) = Thus, we obtain a homomorphism δ from T to T b defining δ () = ν () for ever variable in Q Foundations of Databases 21 Proof: Sufficienc Assume there is a hom δ from T to T. We will show that Q Q We denote the bod of Q as B and the bod of Q as B Let I be a db instance. We show that Q(I) Q (I) Let a Q(I). We show that a Q (I) Since a Q(I), there is a valuation ν : var(q) dom(i) such that ν satisfies B, the bod of Q (that is, ν(b) I in logic programming perspective) ν( ) = a

12 Foundations of Databases 22 Proof: Sufficienc (cntd.) Let ν := ν δ. Then ν is a valuation for Q Moreover, if R( t) is an atom of B, then δ(r( t)) = R(δ( t)) is an atom of B, since δ is a homomorphism, and ν (R( t)) = ν(r(δ( t))) is an atom in I, since ν satisfies B over I, that is, ν (R( t)) is an atom in I In summar, since R( t) was arbitrar, we have that ν (B ) I, that is, ν satisfies B over I Also ν ( ) = ν(δ( )) = ν( ) = a Hence, a Q (I) Foundations of Databases 23 Appling the Homomorphism Theorem: Q 1 Q 2 T1 T2 A B A B f()=, f()= z w f(u)=z, f(w)= u Hence Q1 Q2 T1 T2 A B z A w B f()=, f()= f(z)=u u Hence Q2 Q1

13 Foundations of Databases 24 Quer Containment: Eercise Find all containments and equivalences among the following conjunctive queries: q 1 (,) : r(,), r(,z), r(z,w) q 2 (,) : r(,), r(,z), r(z,u), r(u,w) q 3 (,) : r(,), r(z,u), r(v,w), r(,z), r(,u), r(u,w) q 4 (,) : r(,), r(,3), r(3,z), r(z,w) Foundations of Databases 25 Quer Containment: Compleit Given two conjunctive queries, how hard is it to test whether Q Q? It is eas to transform them into tableau, from either SPJ relational algebra queries, or SQL queries, or rule-based queries. However, a polnomial algorithm for deciding equivalence is unlikel to eist: Theorem. Given two tableau, deciding the eistence of a homomorphism between them is NP-complete. Proof Ideas: Reductions of graph problems like Hamiltonian Path or Clique. Also, reduction of 3SAT. (This one is interesting, because it can be modified to prove that containment is harder for generalisations of conjunctive queries, such as CQs with comparisons or CQs over databases with nulls.) In practice, quer epressions are small, and thus conjunctive quer optimization is nonetheless feasible in polnomial time

14 Foundations of Databases 26 The 3-Colorabilit Problem Instance: A graph G = (V,E) Question: Can G be colored with the three colours {r, g, b } in such a wa that two adjacent vertices have a distinct colour? The 3-colorabilit problem is NP-hard Observation: A graph G is 3-colourable if and onl if there is a graph homomorphism from G to the simple S 3, which consists of of three vertices that are connected to each other. Foundations of Databases 27 Reduction of 3-Colorabilit to Containment Given graph G = (V,E), where V = {v 1,...,v n } and E = {(v il,v jl ) i l < j l, 1 l m } We construct queries Q, Q as follows Q() : e(r,b), e(b,r), e(r,g), e(g,r), e(b,g), e(g,b) Q () : e( i1, j1 ),...,e( im, jm ) where 1,..., n are new variables and there is one atom e( il, jl ) for each edge (v il,v jl ) E Clearl, Q Q if and onl if there is a graph homomorphism from G to S 3

15 Foundations of Databases 28 Minimizing Conjunctive Queries Goal: Given a conjunctive quer Q, find an equivalent conjunctive quer Q with the minimum number of joins. Questions: How man such queries can eist? How different are the? Assumption: We consider onl CQs without equalities and inequalities. We call these queries simple conjunctive queries (SCQs). If nothing else is said in this chapter, CQs are simple CQs Terminolog: If Q( ) : R 1 ( u 1 ),...,R k ( u k ) is a CQ, then Q is a subquer of Q if Q is of the form where 1 i 1 < i 2 <... < i l k. Q ( ) : R i1 ( u i1 ),...,R il ( u il ) Foundations of Databases 29 Minimization: Background Theor Proposition 1. Let q be a SCQ with n atoms and q be an equivalent SCQ with m atoms where m < n. Then there eists a subquer q 0 of q such that q 0 has at most m atoms in the bod and q 0 is equivalent to q. Proposition 2. Let q and q be two equivalent minimal SCQs. Then q and q are identical up to renaming of variables. Conclusions: There is essentiall one minimal version of each SCQ Q. We can obtain it b dropping atoms from Q s bod.

16 Foundations of Databases 30 An Algorithm for Minimizing SCQs Given a conjunctive quer Q, transform it into a tableau T. Minimization algorithm: T := T ; repeat until no change choose a row t in T ; end if there is a homomorphism δ: T T \ { t} then T := T \ { t} Output: The quer Q corresponding to the tableau T Foundations of Databases 31 Questions about the Algorithm Does it terminate? Is Q equivalent to Q? Is Q of minimal length among the queries equivalent to Q?

17 Foundations of Databases 32 Minimizing SPJ/Conjunctive Queries: Eample R with three attributes A,B,C SPJ quer Q = π AB (σ B=4 (R)) π BC (π AB (R) π AC (σ B=4 (R))) Translate into relational calculus (instead of normalizing): ` z1 R(,, z 1 ) = 4 1 ` z2 R( 1,, z 2 ) ` 1 R( 1, 1, z) 1 = 4 Simplif, b substituting the constant, and putting quantifiers forward: 1,z 1,z 2 (R(,4,z 1 ) R( 1,4,z 2 ) R( 1,4,z) = 4) Conjunctive quer: Q(,,z) : R(,4,z 1 ),R( 1,4,z 2 ),R( 1,4,z), = 4 Foundations of Databases 33 Minimizing SPJ/Conjunctive Queries (contd) Tableau T : 4 z z z 4 z Minimization, step 1: Is there a homomorphism from T to 1 4 z z 4 z? Answer: No. For an homomorphism δ, we have δ() = (wh?), thus the image of the first row is not in the smaller tableau.

18 Foundations of Databases 34 Minimizing SPJ/Conjunctive Queries (contd) Step 2: Is T equivalent to 4 z z 4 z? Answer: Yes. Homomorphism δ: δ(z 2 ) = z, all other variables sta the same. The new tableau is not equivalent to 4 z 1 4 z or 1 4 z 4 z Because δ() =, δ(z) = z, and the image of one of the rows is not present. Foundations of Databases 35 Minimizing SPJ/Conjunctive Queries (contd) Minimal tableau: 4 z z 4 z Back to conjunctive quer: Q (,,z) : R(,,z 1 ),R( 1,,z), = 4 An SPJ quer: σ B=4 (π AB (R) π BC (R)) Pushing selections: π AB (σ B=4 (R)) π BC (σ B=4 (R))

19 Foundations of Databases 36 Review of the Journe We started with π AB (σ B=4 (R)) π BC (π AB (R) π AC (σ B=4 (R))) Translated into a conjunctive quer Built a tableau and minimized it Translated back into conjunctive quer and SPJ quer Applied algebraic equivalences and obtained π AB (σ B=4 (R)) π BC (σ B=4 (R)) Savings: one join. Foundations of Databases 37 Minimization of Conjunctive Queries: Multiple Relations We consider again the quer: Q(,): B(,),R(,z),R(,w),R(w, ) The tableau was: A B B: A B R: z w w

20 Foundations of Databases 38 Minimization with Multiple Relations The algorithm is the same as before, but one has to tr rows in different relations. Consider the homomorphism where δ(z) = w, and δ is the identit for all other variables. Appling this to the tableau for Q ields B: A B R: A w B w This can t be further reduced, as for an homomorphism δ, δ()=, δ()=. Thus Q is equivalent to Q (, ) : B(, ), R(, w), R(w, ) One join is eliminated. Foundations of Databases 39 Conjunctive Queries with Equalities and Disequalities Equalit/Disequalit atoms =, = a, z, etc Let T, T be the tableau of the parts of conjunctive queries Q and Q with ordinar relations Sufficienc: Q Q if there eists a homomorphism δ: T T such that for each (dis)equalit atom t 1 θt 2 in Q, we have that δ(t 1 )θδ(t 2 ) is logicall implied b the equalit and disequalit atoms in Q However, eistence of a homomorphims is no more a necessar condition for containment. It holds under certain conditions, though. Note: Deciding whether a set of equalit/disequalit atoms A logicall implies an equalit/disequalit atom is (relativel) eas.

21 Foundations of Databases 40 Queries with Comparisons For queries with comparison atoms s t we have to refine our semantics An ordered domain is a nonempt set D with a linear order (written D ) Let D be fied. Wlog, dom = D. That is, from now on, database instances have constants from D. Real database languages support man domains, ordered and not ordered, b tping relation smbols and admitting onl queries that are well-tped. Foundations of Databases 41 Queries with Comparisons (ctnd) We consider conjunctive queries Q( s) : R,C with tuples of terms in the head whose bodies consist of a set of relational atoms R and a set of comparisons C whose constants are elements of D whose comparisons are interpreted over D The semantics of such queries is defined in a straightforward manner

22 Foundations of Databases 42 Queries with Comparisons: Eamples Q() : P(,w), P(,), R(,u), w 5, 2 Q () : P(,), R(,z), 3 How can we determine containment? Foundations of Databases 43 Quer Homomorphism: Definition Instead of tableau homomorphims, one often defines quer homomorphisms. An homomorphism from Q ( ) : R, C to Q( ) : R, C is a substitution δ such that δ( ) = δ(r ) R C = δ(c ). Here, δ( ), δ(r ) and δ(c ) are the etensions of δ to comple sntactic entities. Note that we view R, R, C, and C as sets of atoms. How should we have to define =?

23 Foundations of Databases 44 Quer Homomorphism: Eample Q () : P(,), R(,z), 3 Q() : P(,w), P(,), R(,u), w 5, 2 An homomorphism from Q to Q is δ() =, δ() =, δ(z) = u. Foundations of Databases 45 Homomorphisms: The General Case Eistence of a homomorphism is not a necessar, but a sufficient condition for containment. Theorem: Let Q, Q be two conjunctive queries, possibl with comparisons and inequalities, such that Q and Q have the same distinguished variables. Then Q Q if there eists a homomorphism from Q to Q.

24 Foundations of Databases 46 Containment of Queries with Comparisons Classical eample: Q () : P(u,v), u v Q() : P(,z), P(z,) Q is contained in Q, but there is no homomorphism from Q to Q. Foundations of Databases 47 Checking Containment of Queries with Comparisons: Idea Q () : P(u,v), u v Q() : P(,z), P(z,) Replace Q with an equivalent union of queries: Q { <z } () : P(,z), P(z,), < z Q { =z } () : P(,z), P(z,), = z Q { >z } () : P(,z), P(z,), > z Check whether Q i Q for all Q i (Case Analsis)

25 Foundations of Databases 48 Linearizations We now make this idea formal. Let D be an ordered domain, D be a set of constants from D, W be a set of variables, and let T := D W denote their union. A linearization of T over D is a set of comparisons L over the terms in T such that for an s, t T eactl one of the following holds: L = D s < t L = D s = t L = D s > t. L partitions the terms into classes such that (i) the terms in each class are equal and (ii) the classes are arranged in a strict linear order Foundations of Databases 49 Linearizations (cntd) Remark: A class of the induced partition contains at most one constant Remark: Whether or not L is a linearization ma depend on the domain. Consider e.g., {1 <, < 2 } A linearization L of T over D is compatible with a set of comparisons C if L C is satisfiable over D

26 Foundations of Databases 50 Linearizations of Conjunctive Queries When checking containment of two queries, we have to consider linearizations that contains the constants of both queries Let Q( s) : R, C be a quer with set of comparisons C ranging over I W be the set of variables occurring in q D be a set of constants from I that comprise the constants of q Then we denote with L D (Q) the set of all linearizations of D W that are compatible with the comparisons C of Q Foundations of Databases 51 Linearizations of Conjunctive Queries (cntd) Let Q be as above. Let L be a linearization of T = D W compatible with C. Note: L defines an equivalence relation on T, where each equivalence class contains at most one constant A substitution φ is canonical for L if it maps all elements in an equivalence class of L to one term of that class if a class contains a constant, then it maps the class to that constant. Then Q L is obtained from Q b means of a canonical substitution φ for L as Q L (φ( s)) : φr, φl, that is, we first replace C with L and then eliminate all equalities b appling φ

27 Foundations of Databases 52 Linearizations of Conjunctive Queries (cntd) Consider Q L (φ( s)) : φr, φl, We call Q L a linearization of q w.r.t. L There ma be more than one linearization of q w.r.t. L, but all linearizations are identical up to renaming of variables Note that φ is a homomorphism from q to Q L Foundations of Databases 53 Linear Epansions A linear epansion of q over D is a famil of queries (Q L ) L LD (Q), where each Q L is a linearization of q w.r.t. L. If Q and D are clear from the contet, or do not matter, we write simpl (Q L ) L.

28 Foundations of Databases 54 Containment of Queries with Comparisons Theorem (Klug 88): If Q, Q are conjunctive queries with comparisons over D with set of constants D (Q L ) L is a linear epansion of Q over D, then: Q Q for ever Q L in (Q L ) L, there is an homomorphism from Q to Q L Theorem (van der Meden 92): Containment with comparisons is Π P 2 -complete. Foundations of Databases 55 Quer Optimization and Functional Dependencies Additional equivalences hold if db instances satisf integrit constraints We consider here functional dependencies Eample: Let R have attributes A,B,C. Assume that I(R) satisfies A B. Then it holds that (π AB (R) π AC (R))(I) = R(I) (We have considered the epressions π AB (R) π AC (R) and R as queries.) Tableau can help with these optimizations! π AB (R) π AC (R) as a conjunctive quer: Q(,,z) : R(,,z 1 ),R(, 1,z)

29 Foundations of Databases 56 Quer Optimization and Functional Dependencies (contd) Tableau: z 1 1 z z Using the FD A B infer = 1 Net, minimize the resulting tableau: z 1 z z z z And this sas that the quer is equivalent to Q (,,z): R(,,z), that is, R This is known as the chase technique Foundations of Databases 57 Quer Optimization and Functional Dependencies (contd) General idea: simplif the tableau using functional dependencies and then minimize. Given: a conjunctive quer Q, and a set of FDs F Algorithm: Step 1. Construct the tableau T for Q Step 2. Appl algorithm CHASE(T,F) Step 3. Minimize output of CHASE(T,F) Step 4. Construct a quer from the tableau produced in Step 3

30 Foundations of Databases 58 The CHASE We assume that all FDs are of the form X A, where A is a single attribute. For simplicit, we also assume that the tableau has onl a single relation. The generalisation is straightforward. for each X A in F do for each t 1, t 2 in T such that t 1.X = t 2.X and t 1.A t 2.A do case t 1.A, t 2.A of one nondistinguished variable replace the nondistinguished variable b the other term one distinguished variable, one distinguished variable or constant replace the distinguished variable b the other term two constants output and STOP end end. Foundations of Databases 59 Quer Optimization and Functional Dependencies: Eample 2 R is over A,B,C; F = {B A } Q = π BC (σ A=4 (R)) π AB (R) Q as a conjunctive quer: Q(,,z) : R(4,,z),R(,,z 1 ) Tableau: 4 z CHASE 4 z minimize 4 z z 1 z 4 z 1 4 z 4 z Final result: Q(,,z) : R(,,z), = 4, that is, σ A=4 (R).

31 Foundations of Databases 60 Quer Optimization and Functional Dependencies: Eample 3 Same R and F ; the quer is: Q = π BC (σ A=4 (R)) π AB (σ A=5 (R)) As a conjunctive quer: Q(,,z) : R(4,,z),R(,, z 1 ), = 5 Tableau: 4 z 5 z 1 CHASE 5 z Final result: (the empt quer) This equivalence does not hold without the FD B A Foundations of Databases 61 Quer Optimization and Functional Dependencies: Eample 4 Sometimes simplifications are quite dramatic Same R, FD is A B, the quer is Q = π AB (R) π A (σ B=4 (R)) π AB (π AC (R) π BC (R)) Convert into conjunctive quer: Q(,) : R(,,z 1 ),R(, 1,z),R( 1,,z),R(,4,z 2 )

32 Foundations of Databases 62 Quer Optimization and Functional Dependencies: Eample 4 (contd) Tableau: z 1 1 z 1 z 4 z 2 CHASE 4 z 1 4 z 1 4 z 4 z 2 minimize 4 z 4 4 Foundations of Databases 63 Quer Optimization and Functional Dependencies: Eample 4 (contd) 4 z is translated into Q(, ) : R(,, z), = 4 4 or, equivalentl π AB (σ B=4 (R)). Thus, π AB (R) π A (σ B=4 (R)) π AB (π AC (R) π BC (R)) = π AB (σ B=4 (R)) in the presence of FD A B. Savings: 3 joins! This cannot be derived b algebraic manipulations, nor conjunctive quer minimization without using CHASE.

33 Foundations of Databases 64 Questions about the CHASE Does the CHASE algorithm terminate? What is the run time? What is the relation between a tableau and its CHASE d version? Quer containment wrt a set of FD s: How can we define this problem? Can we decide this problem? Quer minimsation wrt to a set of FDs Consider SCQs: we know from previous eercises that all such queries are satisfiable. Is the same true if we assume onl database instances that satisf a given set of FDs F?