Normalization

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 Stanford

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

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.

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.

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

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

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

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

FarkasCSCE 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.

FarkasCSCE 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 +

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

FarkasCSCE 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)

FarkasCSCE 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

FarkasCSCE 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)

FarkasCSCE 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!

FarkasCSCE 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.

FarkasCSCE NF 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

FarkasCSCE NF 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)