1. Schema Refinement and Normal Forms Given a design, how do we know it is good or not? What is the best design? Can a bad design be transformed into a good one? Conceptual design SchemasICs

2. Normalization A relation is said to be in a particular normal form if it satisfies a certain set of constraints. If a relation is in a certain normal form (BCNF, 3NF etc.), we know what problems it has and what problems it does not have Each normal eliminates or minimizes certain kinds of problems Given a relation, the process of making it to be in certain normal form is called normalization Typically this is done by breaking up the relation into a set of smaller relations that possess desirable properties.

3. Boyce-Codd Normal Form (BCNF) A relation R is in BCNF if whenever a FD X  A holds in R, one of the following statements is true. X  A is a trivial FD. X is a superkey. A trivial FD X  Y where Y X key nonkey attr_1nonkey attr_nnonkey attr_2 … Example 1: − Scheme: Hourly_Emps (ssn, name, lot, rating, hrly_wages, hrs_worked) − Constraints: ssn is the primary key and Rating  hrly_wages Example 2: −Schema: R(A, B, C, D) −Constraints: A is the primary key, B is a candidate key, is R a BCNF?

4. A BCNF relation does not allow redundancy Every field of every tuple records a piece of information that cannot be inferred from the values in all other non-key fields BCNF is the most desirable form

5. If a relation is not in BCNF, can we make it BCNF? Normalization Example Relation R(SNLRWH) has FDs S  SNLRWH and R  W Second FD causes violation of BCNF –consequence: W values repeatedly associated with R values We decompose SNLRWH into SNLRH and RW

6. A decomposition of a relation schema R The replacement of the schema R by two or more relation schemas, each contains a subset of R and together include all attributes of R. A decomposition must ensure two properties: Lossless join Dependency preservation Normalization through Decomposition

7. Lossless Join Decompositions Decomposition of R into X and Y is a lossless- join decomposition w.r.t. a set of FDs F if, for every instance r that satisfies F: – (r) (r) = r It is always true that r (r) (r) – In general, the other direction does not hold! If it does, the decomposition is lossless-join. It is essential that all decompositions used to deal with redundancy be lossless!

8. A decomposition of D={R 1,R 2,…,R m } of Relation R has the lossless join property with respect to the set of FDs F on R if for every relation instance r(R) that satisfies F, the following holds: NATURAL_JOIN( ) =r

9. Lossless Join Decomposition: Property 1 Property 1: A decomposition D={R1,R2} of R has the lossless join property with respect to a set of FDs F of R if and only if either is in F + or is in F +. Case 1: Case 2: R1 A B R2 B C A foreign key R1 A B R2 B C A foreign key The common attribute must be a super key for either R1 or R2.

10. Lossless Join Decomposition: Property 2 If (i) a decomposition D={R 1,…,R m } of R has the lossless join property with respect to a set of FDs F on R, and (ii) a decomposition D i ={Q 1,…,Q 2 } of R i has the lossless join property with respect to the projection of F on R i. then the decomposition D’={R 1,R 2,…, R i-1,Q 1,…,Q n,R i+1,…,R m } of R has the lossless join property with respect to F. Decomposition D R R 1 R 2 R 3 … R m Decomposition D 3 R3R3 Q 1 Q 2 Q 3 … Q n Decomposition D’ R R 1 R 2 Q 1 Q 2 … Q n R 4 … R m

11. Lossless Join Decomposition into BCNF relations Algorithm: 1 Set D  {R} 2While there is a relation schema Q in D that is not in BCNF do begin Choose a relation schema Q in D that is not in BCNF; Find a functional dependency X  Y in Q that violates BCNF; Replace Q in D by two schemas (Q-Y) and (XUY) end; We have Since X  Y is in F, D={(Q-Y),(X Y)} has the lossless join property.

12. Exercise Determine whether D={R1,R2, R3} of R(S,E,P,N,L,H) is a lossless-join decomposition. R1={S,E} R2={P,N,L} R3={S,P,H} F={S  E, SP  H, P  NL}

13. R with a set of FDs F projection R 1 R 2 R n F R1 F R2 F Rn The projection of F on R i (F Ri ) is defined as: D={R 1,…,R n } of R is dependency preserving with respect to F if. DEPENDENCY PRESERVATION D={R 1,…,R n } is a decomposition of R., meaning that is equivalent to F.

14. We want to preserve the dependencies because each FD in F represents a constraint on the database. We want each original FD to be represented by some individual relation Ri so we can check the constraint without joining two or more relations. Otherwise, each update would require to do join operations Example: Contracts (C S J D P Q V) Contracts(contractid, supplierid, projectid, deptid, partid, qty, value) DEPENDENCY PRESERVATION C  CSJDPQV JP  C SD  P J  S Is it BCNF?

15. Example: Contracts (C S J D P Q V) SDP CSDJQV JS CJDQV SD  P JSJS DEPENDENCY PRESERVATION C  CSJDPQV JP  C SD  P J  S Loss-less join decomposition?

16. Example: Contracts (C S J D P Q V) SDP CSDJQV JS CJDQV SD  P JSJS Where JP  C is in the result of the decomposition? To enforce JP  C, we need to join the three relations for each update DEPENDENCY PRESERVATION C  CSJDPQV JP  C SD  P J  S

17. Example: Contracts (C S J D P Q V) JS CJDPQV J->S Is this decomposition lossless join and dependency preserving? An alternative decomposition C  CSJDPQV JP  C SD  P J  S

18. Question Can any relation be decomposed into BCNF while ensuring lossless join and dependency preservation?

19. In general, there may not be a dependency preserving decomposition into BCNF. – e.g., CSZ, CS  Z, Z  C – what NF? – Let’s consider a decomposition D={ZC,SZ}. – what NF? Is lossless-join decomposition? – Is CS  Z preserved? – CS + ={C,S,Z}; SZ + = {S,Z,C}; ZC + ={Z,C} – R1(ZC); F R1 ={Z  C} – R2(SZ); F R2 ={SZ  Z, SZ  S} = {Z  C,SZ  Z,SZ  S} + Is CS  Z in ?