BCNF分解过程

Question

What is the BCNF decomposition for these dependencies?

A->BCD
BC->DE
B->D
D->A

What is the process to get to the answer?

Answer 1

We can first convert the relation R to 3NF and then to BCNF.

To convert a relation R and a set of functional dependencies( FD's ) into 3NF you can use Bernstein's Synthesis . To apply Bernstein's Synthesis -

• First we make sure the given set of FD's is a minimal cover
• Second we take each FD and make it its own sub-schema.
• Third we try to combine those sub-schemas

For example in your case:

R = {A,B,C,D,E}
FD's = {A->BCD,BC->DE,B->D,D->A}

First we check whether the FD's is a minimal cover ( singleton right-hand side , no extraneous left-hand side attribute, no redundant FD )

• Singleton RHS: We write the FD's with singleton RHS. So now we have FD's as {A->B, A->C, A->D, BC->D, BC->E, B->D, D->A}
• No extraneous LHS attribute: We remove the extraneous LHS attribute C from FD BC->D and BC->E . So now we have FD's as {A->B, A->C, A->D, B->D, B->E, B->D, D->A}
• No redundant FD's: We remove the redundant dependencies. Now FD's are {A->B, A->C, B->D, B->E, D->A}

Second we make each FD its own sub-schema. So now we have - ( the keys for each relation are in bold )

R ₁ ={ A ,B}
R ₂ ={ A ,C}
R ₃ ={ B ,D}
R ₄ ={ B ,E}
R ₅ ={ D ,A}

Third we see if any of the sub-schemas can be combined. We see that R ₁ and R ₂ have the LHS so they can be combined. Similarly R ₃ and R ₄ can be combined. So now we have -

S ₁ = { A ,B,C}
S ₂ = { B ,D,E}
S ₃ = { D ,A}

This is in 3NF . Now to check for BCNF we check if any of these relations (S ₁ ,S ₂ ,S ₃ ) violate the conditions of BCNF ( ie for every functional dependency X->Y the left hand side ( X ) has to be a superkey ) . In this case none of these violate BCNF and hence it is also decomposed to BCNF .