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I have been told a way to obtain lossless join BCNF but I don't know how to calculate candidate keys (also called super key[s] in some cases) and trivial dependencies.

I have been given the following relation:

  • R = (A, B, C, D, E, F)

...where R is relation name, and A, B, C, D, E, F are its attributes.

The given functional dependencies (FDs) are:

  • A B → C D E F
  • C → D
  • E → F

The algorithm says that the first FD is not non-trivial, but here A B is a super key so we don't break original relation based on this.

Moving to the second FD, it's neither non-trivial nor C is candidate key. So, we break the relation:

  • R1 = (A, B, C, E, F) and R2 = (C, D)

Moving to the third FD it's nor non-trivial neither E is candidate key. So, we break the relation again:

  • R11 = (A, B, C, E), R12 = (E, F) and R2 = (C, D)

Hence, we obtained Loss Less BCNF.

But, I always get confused on how to calculate candidate key(s) and see if a FD is non-trivial, although I am quite aware of the definitions1. I also googled and read some documents but still didn't understood this properly. Please help me to figure out candidate key(s), super key(s), non-trivial FD(s) in the easiest way possible so that I never struck with the concept again.

Why do we only see left one's (A B, C, E) for candidate key and not the right one's ( F, D, C D E F) in the three FDs written above?


1 My definitions:

Super key: A set of one or more attributes which taken collectively, allow us to uniquely identify an entity in an entity set.

Candidate Key: A super key for which no proper subset is also a super key.

0

For determining the candidate keys: 1. Determine the attributes not present in the RHS of the FDs 2. Find the closures of those attributes 3. Identify those closures that determine R. These will be your candidate keys

For determining the super keys: Include the candidate keys with all possible combinations of the rest of the attributes

For determining trivial dependency: $\A \supseteq B$, B is trivially dependent upon A. Locate those FDs.

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