1. Normalization

2. FarkasCSCE 5202 Reading Assignments  Database Systems The Complete Book: Chapters 3.6, 3.7, 3.8  Following lecture slides are modified from Jeff Ullman’s slides for Fall 2002 -- Stanford

3. FarkasCSCE 5203 Closures  Closure of attributes (A + ): find keys  Closure of FDs (S + ): projection of FDs to decompositions of schema  Canonical cover: minimize the number of functional dependencies Important for updates

4. FarkasCSCE 5204 Canonical Cover  Combine FDs if possible and eliminate extraneous attributes: Given a set of FDs S, and an FD X  Y Attribute A is extraneous in X if S logically implies (S-{X  Y})  {(X- A)  Y}. Attribute A is extraneous in Y if the set of FDs (S-{X  Y})  {X  (Y-A)} logically implies S.

5. FarkasCSCE 5205 Canonical Cover  No functional dependency in S c contains an extraneous attribute  Each left side of a functional dependency in S c is unique.

6. FarkasCSCE 5206 Example Canonical Cover  Given: 1.A  BC 2.B  C 3.A  B 4.AB  C  Combine 1 and 3 into: A  BC  From 2 and 4: A is extraneous in 4: B  C  C is extraneous in 1: A  B  Result: A  B, B  C

7. FarkasCSCE 5207 Problems of Relational Database Design  Loss of information (lossless-join)  Inability to represent certain information (dependency preservation)  Repetition of information (normal forms)

8. FarkasCSCE 5208 Desirable Properties of Decomposition – Lossless-join  Lossless-Join: Let R be a relation schema, S a set of FDs on R, R1 and R2 a decomposition of R. R1 and R2 form a lossless-join decomposition if at least one of the following functional dependencies are in S + R1  R2  R1 R1  R2  R2

9. FarkasCSCE 5209 Create Lossless-Join Decomposition  Let R be the DB Schema and f: X  Y an FD in S +  Decompose R into two relations R1(X,Y) R2(R-Y)  Continue for each decomposed relation until cannot be decomposed any more

10. FarkasCSCE 52010 Desirable Properties of Decomposition – Dependency Preserving  Dependency Preservation: all dependencies that hold on the original schema should be able to test on individual schemas after decomposition.  Testing dependency preservations on projections is straight forward.

11. FarkasCSCE 52011 Testing dependency preservations on decomposition  Let S be the original set of FDs on R, and S’ the union of FDs on projections R1, R2,…,Rn.  A decomposition is dependency preserving if S’ + =S +

12. FarkasCSCE 52012 Desirable Properties of Decomposition – Avoid Redundancy Owner Name Owner Address Owner PhoneDog NameDog Breed Dog Color Dog Age Viki13 Lake Rd. Irmo, SC 1-803-777-8989PepperG.S.Black3 Viki13 Lake Rd. Irmo, SC 1-803-777-8989BuddyG.S.Brown1 Viki13 Lake Rd. Irmo, SC 1-803-777-8989AlexandraMixGray5 Viki13 Lake Rd. Irmo, SC 1-803-777-8989MissyLabradorBrown5 Pete300 Main St. Columbia, SC 1-803-657-5678HunterVizslaBrown8 Paul99 Apple St Irmo, SC 1-803-899-8524GiantKuvaszWhite4 FD: O.Name  O.Address, O.Phone D.Name, D.Breed  D.Color, D.Age

13. FarkasCSCE 52013 Decomposition and FDs  Functional dependencies: can be used in designing a relational database to remove the undesired properties  Normalization using FDs: Boyce-Codd Normal Form (BCNF) 3 rd Normal Form (3NF)

14. FarkasCSCE 52014 Boyce-Codd Normal Form  A relation is in Boyce-Codd Normal Form if for all FDs X  A in S+ over R at least one of the followings hold: X  A is a trivial FD if X  A is a nontrivial FD then X is a superkey for schema R  Example: R(Name,Breed,Date, Kennel) FD: Name,Breed,Date  Kennel R is in BCNF

15. FarkasCSCE 52015 Decomposition into BCNF  Compute S + (for FDs in S);  if there is a R that is not BCNF then - let X  A a non-trivial FD on R s.t. X  R is not in S + and X  A = ; - Decompose R into: R1(R-A) R2(X,A)

16. FarkasCSCE 52016 BCNF  Not every BCNF decomposition is dependency preserving  Example: FDs: AB  C and C  B Keys: {A,B} and {A,C} C  B is BCNF violation, therefore need to decompose to R1=(AC) and R2=(BC) Decomposition cannot enforce AB  C!

17. FarkasCSCE 52017 Third Normal Form  Modifies BCNF conditions so no need to decompose in this problem situation  An attribute is prime if it is member of any key  X  A violates 3NF if and only if X is not a superkey and also A is not a prime.

18. FarkasCSCE 52018 3NF Conditions  A relation is in 3NF if for all FDs X  A in S + over R at least one of the followings hold: X  A is a trivial FD if X  A is a nontrivial FD then X is a superkey for schema R Each attribute B in A-X is contained in a candidate key for R

19. FarkasCSCE 52019 3NF and BCNF  BCNF gives Lossless-join No-redundancy Dependency preservation not always possible  3NF gives Lossless-join Dependency preservation May have null values (transitive dependencies)