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 | | The cure: 4NF A relation is in 4NF if for every nontrivial mvd x->>y, x is a superkey (any combination of attributes that is a unique ID in the relation (non-minimal) for the relation. |
 | | As usual, the way to get a relation into 4NF is to decompose it, to get the mvds into separate relations: (manager, phone#) (manager, employee) which are now trivial mvds, making up the entire table. |
 | | SOLUTION: Lossless 4NF decomposition: Movie((Producer violates 4NF (you could choose the other multivariable dependency also): D={Movie((Producer,Movie((Star}, CK={Movie, Producer, Star} It is not in 4NF, since in D none of the left hand sides is a superkey. |
| cse.unl.edu /~agnes/multi_own_0.doc (1312 words) |
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