DecompositionUndesirable Properties of Bad DesignHow to avoid ?DecompositionsRelation DecompositionExample of Relation DecompositionLossless Join DecompositionTesting Lossless JoinExample of L-J DecompositionAnother exampleWhy do we preserve the dependency?Dependency Preservation DecompositionTest of Dependency PreservationTest of Dependency Preservation IIDependency Preserving ExampleNon-Dependency Preserving ExampleMinimal RedundancyNormal Form Decompositions ComparisonLossless Check ExampleLossless Check ExampleSlide 21ConclusionReferenceDecompositionBy Yuhung ChenCS157A Section 2October 27 2005Undesirable Properties of Bad DesignRedundancy, resulting in waste of space and complicated updates (inconsistencies.)Inability to represent certain information – ex Null values.Loss of information.How to avoid ? Properties of information repetition and null values suggest -- Decomposition of relation schema.Properties of information loss -- Non-lossy-join decomposition.DecompositionsThere are careless, “bad” decompositions.There are three desirable properties: 1. Lossless.2. Dependency preservation.3. Minimal redundancy.Relation DecompositionOne of the properties of bad design suggests to decompose a relation into smaller relations.Must achieve lossless-join decomposition.Example of Relation DecompositionLossless Join DecompositionDefinition: Let { R1, R2 } be a decomposition of R (R1 U R2 = R); the decomposition is lossless if for every legal instance r of R:r = ΠR1(r) |X| ΠR2(r)Testing Lossless JoinLossless join property is necessary if the decomposed relation is to be recovered from its decomposition.Let R be a schema and F be a set of FD’s on R, and α = (R1, R2) be a decomposition of R. Then α has a lossless join with respect to F iff R1 ∩ R2 -> R1 (or R1 - R2 ) or R2 ∩ R1 -> R2 (or R2 - R1 ) where such FD exist in Closure of FPS This is a sufficient condition, but not a necessary condition.Example of L-J DecompositionFrom the previous example : R = (ABC) F = {A -> B}R1 = (AB), R2 = (AC)R1∩ R2 = A, R1- R2 = Bcheck A -> B in F ? Yes. Therefore lossless R1 = (AB), R2 = (BC)R1∩ R2 = B, R1 - R2 = A , R2 - R1= Ccheck B -> A in F ? NO check B -> C in F ? NO So, this is lossy join.Another exampleR = (City, Street, Zip) F = {CS -> Z, Z -> C}R1 = (CZ) R2 = (SZ)R1 ∩ R2 = Z , R1 – R2 = (SZ)check Z -> C in F ? YesTherefore, the decomposition to be (CZ) (SZ) is lossless join decomposition.Why do we preserve the dependency?We would like to check easily that updates to the database do not result in illegal relations being created.It would be nice if our design allowed us to check updates without having to compute natural joins.Dependency Preservation DecompositionDefinition: Each FD specified in F either appears directly in one of the relations in the decomposition, or be inferred from FDs that appear in some relation.Test of Dependency PreservationIf a decomposition is not dependency-preserving, some dependency is lost in the decomposition.One way to verify that a dependency is not lost is to take joins of two or more relations in the decomposition to get a relation that contains all of the attributes in the dependency under consideration and then check that the dependency holds on the result of the joins.Test of Dependency Preservation IIFind F - F', the functional dependencies not checkable in one relation.See whether this set is obtainable from F' by using Armstrong's Axioms.This should take a great deal less work, as we have (usually) just a few functional dependencies to work on.Dependency Preserving ExampleConsider relation ABCD, with FD’s :A ->B, B ->C, C ->DDecompose into two relations: ABC and CD.ABC supports the FD’s A->B, B->C.CD supports the FD C->D.All the original dependencies are preserved.Non-Dependency Preserving ExampleConsider relation ABCD, with FD’s: A ->B, B ->C, C->DDecompose into two relations: ACD and BC.ACD supports the FD B ->C and implied FD A ->C.BC supports the FD B->C.However, no relation supports A ->B.So the dependency is not preserved.Minimal RedundancyIn order to achieve the lack of redundancy, we do some decomposition which is represented by several normal forms.Normal Form Decompositions Comparison 3NF Decomposition:lossless.Dependency preserving.BCNF Decomposition:Lossless.Not necessarily dependency-preserving.Component relations are all BCNF, and thus 3NF.4NF Decomposition:Lossless.Not necessarily are all 4NF, and thus BCNF and 3NF.No decomposition is guaranteed to preserve all multi-value dependencies.Lossless Check ExampleConsider five attributes: ABCDEThree relations: ABC, AD, BDEFD’s: A ->BD, B ->EA B C D EABC a1 a2 a3 b14 b15AD a1 b22 b23 a4 b25BDE b21 a2 b33 a4 a5Lossless Check ExampleLossless Check ExampleConclusionDecompositions should always be lossless.Decompositions should be dependency preserving whenever possible.We have to perform the normal decomposition to make sure we get rid of the minimal redundant
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