BCNF分解算法说明

Question

I looked in Decomposing a relation into BCNF answers and tried it on my homework, but i don't get the correct answers, so i ask for help in BCNF decomposition

Consider R=(ABCDEG) & F={BG->CD, G->A, CD->AE, C->AG, A->D} .
I start pick A->D .
Now i got S=(AD), R'=(ABCEG).
I pick G->A .
Now i got S=(AD,AG) R'=(BCEG) .
I pick C->G . Now i think i need to get S=(AD,AG,CG) and R'=(BCE) , But the answer in the end is (AD,AG,CGE,BC) .what went wrong? or perhaps, a better algorithm?

Answer 1

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,G}
FD's = {BG->CD,G->A,CD->AE,C->AG,A->D}

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: So we can write our FD's as { BG->C, BG->D, G->A, CD->A, CD->E, C->A, C->G, A->D}.
• No extraneous LHS attribute: We can remove D from LHS of CD->A and CD->E since D is an extraneous attribute here (As we can get D from C since C->A and A->D ). So we now have {BG->C, BG->D, G->A, C->A, C->E, C->G, A->D}
• No redundant FD's: We note that there are a lot of redundant dependencies here. Removing them we have {BG->C, G->A, C->E, C->G, A->D}

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

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

Third we see if any of the sub-schemas can be combined. We see that R ₃ and R ₄ can be combined as they have the same key. So now we have -

S ₁ = {B,G,C}
S ₂ = {A,G}
S ₃ = {C,E,G}
S ₄ = {A,D}

This is in 3NF . Now to check for BCNF we check if any of these relations (S ₁ ,S ₂ ,S ₃ ,S ₄ ) 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 .

Note My final answer above is (AD,AG,CGE,BCG) . The solution you expect is (AD,AG,CGE,BC) but that's wrong. The last relation here (S ₁ ) should also have the G attribute as shown above.

Answer 2

Give input: A relation R0 with set (Minimal) of FD's S0.

Output : A decomposition of R into a collection of relations, all of which are in BCNF

Algo: R <- R0, and S <- S0 Repeat till R is in BCNF. If there is a FD X -> Y that violates BCNF condition. Compute {X}+ , and choose {X}+ as one relation as R1, and another R2 as {(R - X + ) UX} Map FD set S on R1 and R2 (determine FDs on R1 and R2). Recursively repeat the algo on R1 and R2.

Rule: 1.Should be attribute preserving. 2.Should be lossless 3.Should be FD preserving

Example: R(xyz) FD xy -> z; key : xy z-> y; FD xy -> z; key : xy z-> y;

Solve: z-> y violet the BCNF condition.

So decompose relation R {z}+= yz; R1(yz) where key is z and R2(xz) key is x {z}+= yz; R1(yz) where key is z and R2(xz) key is x