Next step in progression from 3NF to BCNF

Question

I am working with relation H that is detailed below:

H = {A,B,C,D}

It shows the following functional dependencies (FD[s]):

F = { A -> B, B -> C, {C,D} -> A}

The candidate keys are

{A,D}, {B,D}, {C,D}

I belive relation H is in 3NF.

I also believe there are two FD that prevent this relation from satisfying BCNF:

A -> B and B -> C

Then I have:

A -> B 
H1 = {A,B}  F1 = {A -> B} <- this is BCNF
H2 = {A,C,D} F2 = {A -> C, {C,D} -> A} <- this isnt' BCNF

B -> C
H11 = {B,C} F11 = {B -> C} <- this is BCNF
H12 = {A,B,D} F12 = {A -> B} <- this is in BCNF

The questions

I have partial information about this algorithm, so I am not sure about certain things:

1. What should I do with H2? Just throw it away and say that H1, H11, H12 are decompositions of H to meet BCFN?

2. Can this relation be converted to BCNF with functional dependency preservation? (I would say no! because it is impossible to derive the FD {C,D} -> A from F1 sum F11 sum F12. Can you confirm that?

Answer 1

First, you assumption about the 3NF is correct.

Then, in the analysis algorithm to find the BNCF, when you start to remove a dependency X → Y since it violates the BCNF, you should put in the first relation H1 not only XY, but X+, while in the second relation you should have H2 = H - X+ + X.

So, in the first step, the two resulting relations are:

H1 = {A,B,C) with dependencies A → B and B → C and key `A`,
H2 = {A, D} (with no dependencies)

In H1 the dependency B → C violates the BCNF since B is not a superkey, and you can decompose it in:

H11 = {A, B} with the dependency A → B and key A
H12 = {B, C} with the dependency B → C and key B

So, at the end, the final decomposition is H11, H12 and H2. Note that this decomposition determines the loss of the dependency {C, D} → A, and this happens also if we start by considering first the dependency B → C (obtaining the same decomposition).