The key point about armstrongs axioms is that they

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The key point about Armstrong’s axioms is that they are bothsoundandcomplete. That is, if we start with a setFof FDs then:Repeated application of Armstrong’s axioms toFgenerates only FDsthat are consequences ofFAny FD that is a consequence ofFbe obtained by repeated applicationof Armstrong’s axioms toF.Jianxin Li (UWA)Redundancy28 / 41
ExampleConsider a relation with attributesABCand letF={AB,BC}Then from transitivity we getAC, by augmentation we getACBCandby union we getABC.FDs that arise from reflexivity such asABBare known astrivialdependencies.Jianxin Li (UWA)Redundancy29 / 41
ClosureGiven a setXof attributes from some relationR, theclosureX+is the set ofall attributes that are determined byX. In symbolsX+={A:XA}The following properties hold:XX+X+=Rif and only ifXis a superkeyXX+is a sort of “maximal” FDJianxin Li (UWA)Redundancy30 / 41
ExampleSupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBDE+? (We useBDEas shorthand for{B,D,E}.)31 / 41
ExampleSupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBDE+? (We useBDEas shorthand for{B,D,E}.)31 / 41
ExampleSupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBDE+? (We useBDEas shorthand for{B,D,E}.)31 / 41
ExampleSupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBDE+? (We useBDEas shorthand for{B,D,E}.)
ExampleSupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBDE+? (We useBDEas shorthand for{B,D,E}.)
Example 2SupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBD+?Jianxin Li (UWA)Redundancy32 / 41
Example 2SupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBD+?Jianxin Li (UWA)Redundancy32 / 41
Example 2SupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBD+?So far we know thatBD+containsBDJianxin Li (UWA)Redundancy32 / 41
Example 2SupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBD+?So far we know thatBD+containsBDFrom the FDDC,BD+containsBCDJianxin Li (UWA)Redundancy32 / 41
Example 2SupposeR(A,B,C,D,E)has the following FDsDC,CEA,DA,AEDWhat isBD+?So far we know thatBD+containsBDFrom the FDDC,BD+containsBCDFrom the FDDA,BD+containsABCDThereforeBD+is not a superkey becauseEis not determined by it.

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Relational model, Database normalization, Jianxin Li

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