DataBase - Is this relation R is in BCNF and dependency preserving?

Question

For R(A,B,C,D,E,G,H) here's the minimal cover:

{A->E,D->H,D->G,E->C,G->B,G->C,H->D}

Candidate keys:

{AH,AD}

By the definition of BCNF, none of the attributes on left side are SK or CK. Thus, it's not in BCNF. Is it safe to conclude that all of the FDs are violating BCNF? If so, in the process of decomposition to BCNF, as the algorithm says, to take the FD that violates BCNF, for example: X->Y, and do the procedure of R1(XY) and R2(R-Y)

In our case, do I need to that do that all over the FDs? If I do so, I get in the end

R1(AE), R2(EC), R3(GB), R4(DH), R5(DG) and R6(AD) 

But still missing G->C and H->D and R6 isn't in the FD from the start. So that doesn't make it dependency preserving?

Answer 1

Is it safe to conclude that all of the FDs are violating BCNF?

Yes

... and do the procedure of R1(XY) and R2(R-Y)

The standard analysis algorithm decompose the original schema in two subschemes R1(X+), R2(R - (X+ - X)). So if you start, for example, from AE, you produce R1(AEC) (since A+ = AEC), and R2(ABDGH). Then you repeat the steps in the remaining relations, if there are other dependencies that violates the BCNF.

For instance, in this case a decomposition that can be obtained is:

R4(AH)
R5(BG)
R6(DGH)
R7(CE)
R8(AE)  

Note that, with this decomposition, the dependency G -> C is not preserved (it is a known fact that the algorithm can produce the loss of one or more dependencies).