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databaserelational-databasebcnf

Boyce-Codd Normal Form

I do not understand Boyce-Codd Normal Form. I looked at my textbook but I do not understand it. Let's say relation R = (A,B,C,D,E) and F={A->C, D->CB, AC->E}. How do you determine if R is in BCNF? Need help. Thanks!

Solution

  • To determine if a relation is BCNF we examine it's functional dependencies.

    It is in BCNF if for each FD X→Y, we either have

    • X→Y is a trivial functional dependency (Y ⊆ X)
    • X is a superkey for schema R.

    The FDs are A→C, D→CB, AC→E. Let's start with the first FD A→C.

    A→C is not trivial because C ∉ A. A→A is trivial dependency for instance.

    Now is A→C a superkey? To check that we compute the closure of left hand side of the FD, in this case A. The closure is all elements logically implied by A. [A]+ = A ∪ C ∪ E = ACE or so we have A→ACE.

    ACE is not a superkey, because it does contain all attributes of the relation.

    So the relation is not in BCNF, because A→C is neither trivial or a superkey.

    There are other violations of BCNF too. [D]+ = BCD which is not a superkey or trivial. [AC]+ = ACE which is not a superkey or trivial.

    Hope this helps! I think everything is correct but I'm studying for finals right now and trying to learn a lot of this material as well.